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# Linear Algebra MTH 2311

Baylor University

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This 277 page Class Notes was uploaded by Dejon Bednar on Saturday October 3, 2015. The Class Notes belongs to MTH 2311 at Baylor University taught by Ronald Morgan in Fall. Since its upload, it has received 34 views. For similar materials see /class/217913/mth-2311-baylor-university in Mathematics (M) at Baylor University.

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11 SOLUTIONS Linear Equations in Linear Algebra Notes The key exercises are 7 or 11 or 12 19 22 and 25 For brevity the symbols R1 R2 stand for row 1 or equation 1 row 2 or equation 2 and so on Additional notes are at the end of the section H N 2x1 7x2 2 5 5x17x2 11 x1 5x2 2 7 l 5 7 2 7 5 Replace R2 by R2 2R1 and obtain Scale R2 by 13 Replace R1 by R1 5R2 The solution is x1 x2 8 3 or simply 8 3 2x14x2 4 2 4 4 5 7 11 Scale R1 by 12 and obtain Replace R2 by R2 5R1 Scale R2 by 13 Replace R1 by R1 2R2 The solution is x1 x2 12 7 or simply 12 7 x15x2 7 3x2 2 x15x2 7 x2 x1 2 8 x2 3 x12x2 2 2 5x17x2 11 x12x2 2 2 3x2 2 21 x12x2 2 2 x2 2 7 x1 12 x2 2 7 Ulh i O i Or A l Il Il Il I O i t UIUJUI t O t O 2 CHAPTER 1 Linear Equations in Linear Algebra 3 The point of intersection satis es the system of two linear equations x1 5x2 7 1 5 7 x1 2x2 2 1 2 2 5 7 1 5 7 Replace R2 by R2 71R1 and obtain x1 x2 7x2 2 9 0 7 9 x1 5x2 7 1 5 7 Scale R2 by 717 x2 97 0 1 9 7 x1 47 1 0 47 Replace R1 by R1 75R2 x2 9 7 0 1 9 7 The point of intersection is x1 x2 47 97 4 The point of intersection satis es the system of two linear equations x1 5x2 1 1 5 1 3x1 7x25 3 7 5 5 1 1 5 1 Replace R2 by R2 73R1 and obtain x1 x2 8x2 2 2 0 8 2 x1 5x2 1 1 5 1 Scale R2 by 18 x2 14 0 1 1 4 x1 9 4 1 0 9 4 Replace R1 by R1 5R2 x2 1 4 0 1 1 4 The point of intersection is x1 x2 9 4 14 Equot The system is already in Triangular form The fourth equation is x4 75 and the other equations do not contain the variable x The next two steps should be to use the variable x3 in the third equation to eliminate that variable from the rst two equations In matrix notation that means to replace R2 by its sum with 3 times R3 and then replace R1 by its sum with 75 times R3 9 One more step will put the system in triangular form Replace R4 by its sum with 73 times R3 which 1 0 0 2 7 0 4 produces 0 0 1 2 3 After that the next step is to scale the fourth row by 71 5 0 0 0 5 15 gt1 Ordinarily the next step would be to interchange R3 and R4 to put a 1 in the third row and third column But in this case the third row of the augmented matrix corresponds to the equation 0 x1 0 x2 0 x3 1 or simply 0 1 A system containing this condition has no solution Further row operations are unnecessary once an equation such as 0 1 is evident The solution set is empty 11 Solutions 3 8 The standard row operations are 1 4 9 0 1 4 9 0 1 4 0 0 1 0 0 0 0 1 7 0 N 0 1 7 0 N 0 1 0 0 N 0 1 0 0 0 0 2 0 0 0 1 0 0 0 1 0 0 0 1 0 The solution set contains one solution 0 0 0 9 The system has already been reduced to triangular form Begin by scaling the fourth row by 12 and then replacing R3 by R3 3R4 1 1 4 1 1 0 0 4 1 100 4 01 30 701 30701 30 7 0 0 l 3 l 0 0 l 3 l 0 0 l 0 5 0 0 0 4 0 0 0 l 2 0 0 0 l 2 Next replace R2 by R2 3R3 Finally replace R1 by R1 R2 1 l 0 0 4 l 0 0 0 0 l 0 0 8 0 l 0 0 8 N 0 0 1 0 5 N 0 0 1 0 5 0 0 0 l 2 0 0 0 l 2 O O N The solution set contains one solution 4 8 5 2 10 The system has already been reduced to triangular form Use the l in the fourth row to change the 41 and 3 above it to zeros That is replace R2 by R2 4R4 and replace R1 by R1 N3R4 For the nal step replace R1 by R1 2R2 1 203 21 20071000 3 010 470100 5 0100 5 00106N00106N00106 0001 30001 30001 3 The solution set contains one solution N3 N5 6 N3 11 First swap R1 and R2 Then replace R3 by R3 N3Rl Finally replace R3 by R3 2R2 014 5135 2135 2135 2 135 2N014 5N014 5N014 5 377637760 2 8120002 The system is inconsistent because the last row would require that 0 2 if there were a solution The solution set is empty 12 Replace R2 by R2 N3Rl and replace R3 by R3 4R1 Finally replace R3 by R3 3R2 1 3 4 4 1 3 4 4 1 3 4 4 3 7 7 8N0 2 5 4N0 2 5 4 4 6 1 7 0 6 15 9 0 0 0 3 The system is inconsistent because the last row would require that 0 3 if there were a solution The solution set is empty 4 CHAPTER 1 Linear Equations in Linear Algebra 1 0 3 8 1 0 3 8 1 0 3 8 1 0 3 8 1322 9 7N0215 9N015 2N015 2 0 1 5 2 0 1 5 2 0 2 15 9 0 0 5 5 1 0 3 8 1 0 0 5 N 0 1 5 2 N 0 1 0 3 The solution is 5 3N1 0 0 1 1 l 0 0 1 1 3 0 5 1 3 0 5 1 3 0 5 1 3 0 5 14 1 52N0 257N0 110N0110 0 1 l 0 0 1 1 0 0 2 5 7 0 0 7 7 1 3 0 5 l 3 0 5 1 0 0 2 N 0 1 1 0 N 0 1 0 1 N 0 l 0 1 The solution is 2N1 1 0 0 1 1 0 0 1 1 0 0 1 1 103 0 2 10 3 0 2 010 3 3 010 3 3 0 23 21N0 2 3 2 1 3 00 7 5 0 0 9 7 11 10 3 0 2 103 0 2 010 3 3 010 3 3 N00 3 4 7N003 47 00 97 11000 510 The resulting triangular system indicates that a solution exists In fact using the argument from Example 2 one can see that the solution is unlque 16 First replace R4 by R4 2Rl and replace R4 by R4 N32R2 One could also scale R2 before adding to R4 but the arithmetic is rather easy keeping R2 unchanged Finally replace R4 by R4 R3 100 2 3100 2 3 220002200 0013 700131 23215032 3 1 100 2 3100 2 3 0220002200 N0013 N00131 00 1 3 100000 The system is now in triangular form and has a solution The next section discusses how to continue with this type of system i n l i n on i 0 N O N p A N N 11 Solutions 5 Row reduce the augmented matrix corresponding to the given system of three equations 1 4 l l 4 1 1 4 1 2 1 3 N 0 7 5 N 0 7 5 1 3 4 0 7 5 0 0 0 The system is consistent and using the argument from Example 2 there is only one solution So the three lines have only one point in common Row reduce the augmented matrix corresponding to the given system of three equations 1 2 1 4 1 2 1 4 1 2 1 4 0 1 1 1 N 0 1 1 l N 0 1 1 1 1 3 0 0 0 1 1 4 0 0 0 5 The third equation 0 N5 shows that the system is inconsistent so the three planes have no point in common 1 h 4 1 h 4 N Write c for 6 N 3h Ifc 0 that is ifh 2 then the system has no 3 6 8 0 6 3h 4 solution because 0 cannot equal 41 Otherwise when h 2 the system has a solution 1 h 3 1 h 3 N Write c for 4 2h Then the second equation 0x2 0 has a solution 2 4 6 0 4 211 0 for every value of c So the system is consistent for all h 1 3 2 1 3 2 N Write c for h 12 Then the second equation 0x2 0 has a solution 4 h 8 0 h 12 0 for every value of 0 So the system is consistent for all h 2 3 h 2 3 6 9 5 0 0 ifh 53 h 5 3h The system is consistent if and only if 5 3h 0 that is ifand only a True See the remarks following the box titled Elementary Row Operations b False A 5 X 6 matrix has ve rows P False The description given applied to a single solution The solution set consists of all possible solutions Only in special cases does the solution set consist of exactly one solution lVIark a statement True only if the statement is always true P True See the box before Example 2 a True See the box preceding the subsection titled Existence and Uniqueness Questions 6quot False The de nition of raw equivalent requires that there exist a sequence of row operations that transforms one matrix into the other P False By de nition an inconsistent system has no solution P True This de nition of equivalent systems is in the second paragraph after equation 2 6 CHAPTERI Linear EquationsinLinearAlgebra 1 4 7 g 1 4 7 g 1 4 7 g 25 0 3 5 h N 0 3 5 h N 0 3 5 h 2 5 9 k 0 3 5 k2g 0 0 0 k2gh N 5 N l N on N O M O M p A M N 03 93 Let b denote the number k 2g h Then the third equation represented by the augmented matrix above is 0 b This equation is possible if and only if b is zero So the original system has a solution if and only if k 2g h 0 A basic principle of this section is that row operations do not affect the solution set of a linear system Begin with a simple augmented matrix for which the solution is obviously N2 1 0 and then perform any elementary row operations to produce other augmented matrices Here are three examples The fact that they are all row equivalent proves that they all have the solution set N2 1 0 1 0 0 N2 1 0 0 N2 1 0 0 N2 0 1 0 l N 2 1 0 3 N 2 1 0 N3 0 0 1 0 0 0 1 0 2 0 1 N4 Study the augmented matrix for the given system replacing R2 by R2 wRl 13 f 1 3 f L d M0 Ml This shows that shows d N 30 must be nonzero since f and g are arbitrary Otherwise for some choices of f and g the second row would correspond to an equation of the form 0 b where b is nonzero Thus d 30 Row reduce the augmented matrix for the given system Scale the rst row by 111 which is possible since a is nonzero Then replace R2 by R2 wRl a b f l ba fa l ba fa c d g c d g 0 d cba g cfa The quantity d N c ba must be nonzero in order for the system to be consistent when the quantity g N c f a is nonzero which can certainly happen The condition that d N Cba 0 can also be written as adNbc 0 orad bc Swap R1 and R2 swap R1 and R2 Multiply R2 by Nl2 multiply R2 by N2 Replace R3 by R3 4Rl39 replace R3 by R3 4Rl Replace R3 by R3 3R2 replace R3 by R3 N3R2 The rst equation was given The others are T2T12040T34 or 4T2 T1 T3260 T3T4T240304 or 4T3 T4 T270 T410T1T3304 or 4T4 T1 T340 11 Solutions Rearranging 4T1 T2 T 4 30 T1 4T 2 T3 60 T2 4T3 T 4 70 T1 T3 4T 4 40 34 Begin by interchanging R1 and R4 then create zeros in the rst column 4 1 0 1 30 1 0 1 4 40 1 0 1 4 40 1 4 1 0 60 1 4 1 0 60 0 4 0 4 20 0 1 4 1 70 N 0 1 4 1 70 N 0 1 4 1 70 1 0 1 4 40 4 1 0 1 30 0 1 4 15 190 Scale R1 by 71 and R2 by 14 create zeros in the second column and replace R4 by R4 R3 7 1 0 1 4 40 1 0 1 4 40 1 0 1 4 40 0 1 0 1 5 0 1 0 1 5 0 1 0 1 5 0 1 4 1 70 N 0 0 4 2 75 N 0 0 4 2 75 0 1 4 15 190 0 0 4 14 195 0 0 0 12 270 Scale R4 by 112 use R4 to create zeros in column 4 and then scale R3 by 14 7 1 0 1 4 40 1 0 1 0 50 1 0 1 0 50 0 1 0 1 5 0 1 0 0 275 0 1 0 0 275 0 0 4 2 75 0 0 4 0 120 0 0 1 0 30 0 0 0 1 225 0 0 0 1 225 0 0 0 1 225 The last step is to replace R1 by R1 71R3 1 0 0 0 200 0 1 0 0 275 N The solutlon 1s 20 275 30 225 0 0 1 0 300 0 0 0 1 225 Notes The Study Guide includes a lVIathematical Note about statements If then This early in the course students typically use single row operations to reduce a matrix As a result even the small grid for Exercise 34 leads to about 25 multiplications or additions not counting operations with zero This exercise should give students an appreciation for matrix programs such as MATLAB Exercise 14 in Section 110 returns to this problem and states the solution in case students have not already solved the system of equations Exercise 31 in Section 25 uses this same type of problem in connection with an LU factorization For instructors who wish to use technology in the course the Study Guide provides boxed MATLAB notes at the ends of many sections Parallel notes for Maple Mathematica and the TI838689 and HP48G calculators appear in separate appendices at the end of the Study Guide The MATLAB box for Section 11 describes how to access the data that is available for all numerical exercises in the text This feature has the ability to save students time if they regularly have their matrix program at hand when studying linear algebra The MATLAB box also explains the basic commands replace swap and scale These commands are included in the text data sets available from the text web site www1aylinalgebracom 8 CHAPTER 1 Linear Equations in Linear Algebra 12 SOLUTIONS Notes The key exercises are 1720 and 23728 Students should work at least four or ve from Exercises 7714 in preparation for Section 15 1 Reduced echelon form a and b Echelon form d Not echelon c 2 Reduced echelon form a Echelon form b and d Not echelon c 1234 2341234 34 670 3 6 90 1 2 3 6 8 9 0 5 10 15 0 5 10 15 1234 0 1 2 234 01230 2 3Pivotcols1and24 67 00000000 6789 1357135713571357 435790 4 8 1201230123 5791 8 16 34 0 8 16 34 00 0 10 13571350 0 10 357 Pivotcols 012301200 20152and4 3 79 00010001000 579 II0I II0I 5 60l0000 00000 000000 7134713471347 30 5 39397600 5 15001300 3 3 Corresponding system of equations x2 The basic variables corresponding to the pivot positions are x1 and x3 The remaining variable x2 is free Solve for the basic variables in terms of the free variable The general solution is x1 5 3x2 x2 is free 263 Note Exercise 7 is paired with Exercise 10 12 Solutions 9 8 1 4 0 7 1 4 0 7 1 4 0 7 D 0 0 9 I 2 7 0 10 0 l 0 4 0 1 0 4 0 0 4 9 Corresponding system of equatlons 4 The basic variables corresponding to the pivot positions are x1 and xi The remaining variable 03 is free Solve for the basic variables in terms of the free variable In this particular problem the basic variables do not depend on the value of the free variable 9 9 General solution x2 4 x3 is free Note A common error in Exercise 8 is to assume that x3 is zero To avoid this identify the basic variables rst Any remaining variables are free This type of computation will arise in Chapter 5 901 651 27 6 0 54 391 27 6 01 65 0 65 5 4 6x35 Corresponding system x145x3 Basic variables x1 x2 free variable x3 General solution x2 5 6x3 x3 is free 101 2 131 2 13 20 4 393 6 22 0 01 7 0 0 7 2 4 Corresponding system x2 7 x1 4 2x2 Basic variables x1 x3 free variable x2 General solution x2 is free 3 4 2 0 3 4 2 0 CD 43 23 0 11 9 12 6 0 N 0 0 0 0 N 0 0 0 0 6 8 4 0 0 0 0 0 0 0 0 0 Corresponding system 0 0 10 CHAPTERI Linear Equations in Linear Algebra 4 2 x x 3 2 3 3 Basic variable x1 free variables x2 x3 General solution x2 is free x3 is free 1 7 0 6 5 1 7 0 6 5 7 0 6 5 12 0 0 1 2 3 N 0 0 1 2 3 N 0 0 G 2 3 1 7 4 2 7 0 0 4 8 12 0 0 0 0 0 7x2 6x4 5 Corresponding system 9 2x4 3 0 0 x1 5 7x2 6x4 x2 is free Basic variables x1 and 0 free variables x2 x4 General solution x3 3 2x4 x4 is free 1 30 10 21 30092 000 35 130100 41N0100 41NOCD00 41 I 0 0 0 l 9 4 0 0 0 1 9 4 0 0 0 9 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3x5 5 Corresponding system 4x5 2 1 9x5 4 0 0 x153x5 xz14x5 Basic variables x1 x2 x4 free variables x3 x5 General solution x3 is free x44 9x5 x5 is free Note The Study Guide discusses the common mistake x3 0 12 5 60 5 Do 7 00 9 1401 6 30 2N0 6 30 2 3900 0 010 00 0 0 0 00 0 00 0 00 0 00 0 p A UI gt1 9 0 2 O 21 12 Solutions 11 72 9 6 3x 2 Corresponding system g 4 0 0 0 x1 9 72 x2 2 62 3x4 Basic variables x1 x2 x5 free variables x3 6 General solution x3 is free x4 is free x5 0 a The system is consistent with a unique solution b The system is inconsistent The rightmost column of the augmented matrix is a pivot column a The system is consistent with a unique solution b The system is consistent There are many solutions because 62 is a free variable 2 3 h 3 3 h N The system has a solutlon only 1f7 7 2h 0 that 1s 1fh 72 4 6 7 0 0 7 2h 1 3 5 11 because 0 cannot equal 3 Otherwise when 11 15 the system has a solution 1 h 2 D h 2 4 8 k 0 8 411 k 8 a When 11 2 and k 8 the augmented column is a pivot column and the system is inconsistent 2 3 2 N Ifh 15 1s zero that 1s 1fh 715 then the system has no solutlon 7 0 11 15 3 b When 11 2 the system is consistent and has a unique solution There are no free variables c When 11 2 and k 8 the system is consistent and has many solutions 1 3 2 D 3 2 3 h klNio 11 9 k 6 a When 11 9 and k 6 the system is inconsistent because the augmented column is a pivot column b When h 9 the system is consistent and has a unique solution There are no free variables P When 11 9 and k 6 the system is consistent and has many solutions False See Theorem 1 False See the second paragraph of the section True Basic variables are defined after equation 4 True This statement is at the beginning of Parametric Descriptions of Solution Sets False The row shown corresponds to the equation 5x4 0 which does not by itself lead to a contradiction So the system might be consistent or it might be inconsistent F997 to 22 N 03 N J N UI N 5 CHAPTER 1 Linear Equations in Linear Algebra w False See the statement preceding Theorem 1 Only the reduced echelon form is unique 9 False See the beginning of the subsection Pivot Positions The pivot positions in a matrix are determined completely by the positions of the leading entries in the nonzero rows of any echelon form obtained from the matrix P True See the paragraph after Example 3 Q False The existence of at least one solution is not related to the presence or absence of free variables If the system is inconsistent the solution set is empty See the solution of Practice Problem 2 True See the paragraph just before Example 4 5 Yes The system is consistent because with three pivots there must be a pivot in the third bottom row of the coefficient matrix The reduced echelon form cannot contain a row of the form 0 0 0 0 0 1 The system is inconsistent because the pivot in column 5 means that there is a row of the form 0 0 0 0 1 Since the matrix is the augmented matrix for a system Theorem 2 shows that the system has no solution If the coef cient matrix has a pivot position in every row then there is a pivot position in the bottom row and there is no room for a pivot in the augmented column So the system is consistent by Theorem 2 Since there are three pivots one in each row the augmented matrix must reduce to the form 0 0 a a 0 G 0 b andso b 0 0 G c 9 c No matter what the values of a b and c the solution exists and is unique If a linear system is consistent then the solution is unique and only every column in the coe icient matrix is a pivot column otherwise there are infinitely many solutions This statement is true because the free variables correspond to nonpivot columns of the coefficient matrix The columns are all pivot columns if and only if there are no free variables And there are no free variables if and only if the solution is unique by Theorem 2 Every column in the augmented matrix except the rightmost column is a pivot column and the rightmost column is not a pivot column An underdetermined system always has more variables than equations There cannot be more basic variables than there are equations so there must be at least one free variable Such a variable may be assigned infmitely many different values If the system is consistent each different value of a free variable will produce a different solution x1 x2 g 4 Example 2 2x2 2 5 Yes a system of linear equations with more equations than unknowns can be consistent x1 x2 2 Example in which x1 x2 1 x1 x2 0 3x1 2x2 5 32 According to the numerical note in Section 12 when n 30 the reduction to echelon form takes about 12 Solutions 13 23033 18000 ops while further reduction to reduced echelon form needs at most 302 900 ops Of the total ops the backward phase is about 900 18900 048 or about 5 When n 300 the estimates are 230033 18000000 phase for the reduction to echelon form and 3002 90000 ops for the backward phase The fraction associated with the backward phase is about 9X104 18X106 005 or about 5 33 For a quadratic polynomial pt a0 alt 1212 to exactly t the data 1 12 2 15 and 3 16 the coef cients a0 a1 12 must satisfy the systems of equations given in the text Row reduce the augmented matrix 1 1 1 12 1 1 1 12 1 1 2 4 15 N 0 1 3 3 N 0 1 3 9 16 0 2 8 4 0 1 1 0 13 D 0 0 7 N 0 1 0 6 N 0 D 0 6 0 0 1 1 0 0 D 1 The polynomial is pt 7 6tN 12 34 M The system of equations to be solved is 10 1180 12802 13803 10 1182 12822 13823 10 1184 12842 13843 10 1186 12862 13863 10 1188 12882 13883 10 a1 10 121102 131103 The unknowns are 10 11 matrix 1 0 0 0 0 0 1 2 4 8 16 32 1 4 16 64 256 1024 1 6 36 216 1296 7776 1 8 64 512 4096 32768 1 10 102 103 104 105 1 0 0 0 0 0 0 2 4 8 16 32 0 0 8 48 224 960 N 0 0 0 48 576 4800 0 0 0 192 2688 26880 0 0 0 480 7680 90240 1 1 0 NWD I 12 3 2 141104 0 29 148 396 N 743 OOOOOD I OOOOOD I OOOONO OOOONO 1 1 1 N 0 1 3 0 0 1 24 48 OOOOQJRO a5 1 105 15 Use technology to compute the reduced e 12 3 1 0 290 148 396 743 119 0 16 224 1248 4032 9920 0 32 960 4800 7680 42240 0 16 224 576 384 1920 960 7680 32640 99840 chelon of the augmented 0 32 0 29 309 627 1045 0 29 39 69 245 14 CHAPTERl Linear Equations in Linear Algebra 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 2 4 8 16 32 29 0 2 4 8 16 32 29 0 0 8 48 224 960 9 0 0 8 48 224 960 9 N 0 0 0 48 576 4800 39 0 0 0 48 576 4800 39 0 0 0 0 384 7680 69 0 0 0 0 384 7680 69 70 0 0 0 0 3840 107 0 0 0 0 0 1 0026 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 2 4 8 16 0 28167 0 1 0 0 0 0 17125 0 0 8 48 224 0 65000 0 0 1 0 0 0 11948 N0 0 0 48 576 0 86000 WW0 0 0 1 0 0 6615 0 0 0 0 384 0 26900 0 0 0 0 1 0 0701 70 0 0 0 0 1 0026047 0 0 0 0 0 1 0026 Thus pt 1712517 119482 6615 7 07017 0026 and p75 646 hundred 1b Notes In Exercise 34 if the coef cients are retained to higher accuracy than shown here then p75 648 If a polynomial of lower degree is used the resulting system of equations is overdetermined The augmented matrix for such a system is the same as the one used to nd p except that at least column 6 is missing When the augmented matrix is row reduced the sixth row of the augmented matrix will be entirely zero except for a nonzero entry in the augmented column indicating that no solution exists Exercise 34 requires 25 row operations It should give students an appreciation for higherlevel commands such as gauss and bgauss discussed in Section 14 of the Study Guide The command ref reduced echelon form is available but I recommend postponing that command until Chapter 2 The Study Guide includes a Mathematical Note about the phrase If and only if used in Theorem 2 13 SOLUTIONS Notes The key exercises are 11714 17722 25 and 26 A discussion of Exercise 25 will help students understand the notation a1 a2 213 211 a2 a3 and Spana1 a2 213 H W 1 W 1 u v 2 1 2 1 1 Using the de nitions carefully u 2v I 2 3 1 GEES 14 6 5 or more quickly 2 1 2 2 1 2 2 4 1 3 1 6 5 u 2v 2 The mterrnedrate step 1s often not wr1tten 2 1 2 2 4 2 u31jlzifnllil39 Using the de nitions carefully 9 gt1 13 39 Fl H W ml M W 39 u 2v 2 or more qulckly 2 1 2 2 l 2 2 4 3 2 3 4 1 u 2v 2 The intermediate step is often not written 2 1 2 2 4 Solutions 1 5 1 6x1 3x2 1 x1 1 x2 4 7 x1 4x2 7 x14x2 7 5 0 5 5x1 0 5 5x1 5 6x1 3x2 1 x1 4x2 7 5x1 5 Usually the intermediate steps are not displayed 2 8 1 0 2x1 8x2 g 0 2x1 8x2 g 0 x1 x2 g 3 5 6 0 3x1 5x2 6x3 0 3x1 5x2 6x3 0 2x2 8x2 x3 0 3x1 5x2 6x3 0 Usually the intermediate steps are not displayed See the gure below Since the grid can be extended in every direction the gure suggests that every vector in R2 can be written as a linear combination of u and v To write a vector 21 as a linear combination of u and v imagine walking from the origin to 21 along the grid quotstreetsquot and keep track of how many quotblocksquot you travel in the udirection and how many in the vdirection a To reach a from the origin you might travel 1 unit in the udirection and 72 units in the vdirection that is 2 units in the negative vdirection Hence a u 7 2v 16 CHAPTERI Linear Equations in Linear Algebra b To reach b from the origin travel 2 units in the udirection and 72 units in the vdirection So b 2H 7 2v Or use the fact that b is 1 unit in the udirection from a so that bauu72vu2u72v c The vector 0 is 715 units from b in the vdirection so 0 b715v 2u 72v 7 15v 2u 735v d The map suggests that you can reach d if you travel 3 units in the udirection and 41 units in the vdirection If you prefer to stay on the paths displayed on the map you might travel from the origin to 73v then move 3 units in the udirection and nally move 71 unit in the vdirection So d73v3u7v3u74v Another solution is db72vu2u72v72vu3u74v 8 See the gure above Since the grid can be extended in every direction the gure suggests that every vector in R2 can be written as a linear combination of u and v w To reach w from the origin travel 71 units in the udirection that is 1 unit in the negative udirection and travel 2 units in the vdirection Thus w 71u 2v or w 2v 7 u 94 To reach x from the origin travel 2 units in the vdirection and 72 units in the udirection Thus x 72u 2v Or use the fact that x is 71 units in the udirection from w so that xw7u7u2v7u72u2v The vector y is 15 units from x in the vdirection so y x 15v 72u 2v15v 72u 35V The map suggests that you can reach z if you travel 4 units in the vdirection and 73 units in the udirection So z 4v 7 3H 73u 4v If you prefer to stay on the paths displayed on the map you might travel from the origin to 72u then 4 units in the vdirection and nally move 71 unit in the udirection So z72u4v7u73u4v lt N x2 5X3 0 x2 5x3 0 9 4x1 6x2 g 0 4x1 6x2 x3 0 7x1 3x2 8 0 7x1 3x2 82g 0 0 x2 5x3 0 0 1 5 0 4x16x2 7x30 x1 4x26x3710 7x1 3x2 78x3 0 71 3 78 0 Usually the intermediate calculations are not displayed 13 Solutions 17 Note The Study Guide says Check with your instructor whether you need to show work on a problem such as Exercise 9 4x1 x2 3mg 9 4x1 x2 3x3 10 x1 7x2 2Jg 2 x1 7x2 2Jg 2 8x1 6962 5Jg 15 8x1 6x2 5 15 4x1 x2 3x3 9 4 1 3 9 x1 7x2 2x3 2 x1 1x2 7 Jg 2 2 8x1 6x2 5x3 15 8 6 5 15 Usually the intermediate calculations are not displayed 11 The question Is b a linear combination of a1 a2 and a3 is equivalent to the question Does the vector equation x1211 x2212 x3213 b have a solution The equation 1 0 5 2 x1 2 x2 1g 6 1 0 2 8 6 a1 a2 213 b has the same solution set as the linear system whose augmented matrix is 1 0 5 2 M 2 1 6 1 0 2 8 6 Row reduceM until the pivot positions are visible 1 0 5 2 D 0 5 2 M N 0 1 4 3 N 0 D 4 3 0 2 8 6 0 0 0 0 The linear system corresponding toM has a solution so the vector equation has a solution and therefore b is a linear combination of a1 a2 and a3 12 The equation 1 0 2 5 x1 2 x2 5 x3 0 11 2 5 8 7 a1 a2 213 b has the same solution set as the linear system whose augmented matrix is 18 CHAPTERI Linear Equations in Linear Algebra 1 0 2 5 M 2 5 0 11 2 5 8 7 Row reduceM until the pivot positions are Visible 1 0 2 5 0 2 5 MN 0 5 4 1 N 0 4 1 0 5 4 3 0 0 0 The linear system corresponding to M has no solution so the vector equation has no solution and therefore b is not a linear combination of a1 a2 and a3 13 Denote the columns of A by a1 a2 213 To determine if b is a linear combination of these columns use the boxed fact on page 34 Row reduced the augmented matrix until you reach echelon form 1 4 2 3 4 2 3 0 3 5 7 N 0 5 7 2 8 4 3 0 0 0 The system for this augmented matrix is inconsistent so b is not a linear combination of the columns of A 1 2 6 11 D 2 6 11 14 a1 a2 213 b 0 3 7 5 N 0 7 5 The linear system correspondingto this 1 2 5 9 0 0 2 matrix has a solution so b is a linear combination of the columns ofA 15 Noninteger weights are acceptable of course but some simple choices are 0v1 0v2 0 and 7 5 1v1 0v2 1 0v1 1v2 3 6 0 12 1v1 1v2 4 1v1 7 1v2 2 6 7 6 16 Some liker choices are 0v1 0v2 0 and 3 2 1v1 0v2 0v1 1v2 O 1v1 1v2 1v1 7 1v2 0 13 Solutions 19 1 2 4 1 2 4 1 2 4 2 4 17 a1 212 b 4 3 1 N 0 5 15 N 0 1 3 N 0 D 3 Thevectorbis 2 7 h 0 3 h8 0 3 h8 0 0 h17 in Spana1 212 when 11 17 is zero that is when 11 N17 1 3 h 1 3 h C 3 h 18 v1 v2 y 0 1 5 N 0 1 5 N 0 5 The vector y is in 2 8 3 0 2 32h 0 0 72h Spanv1 v2 when 7 211 is zero that is when 11 N72 19 By inspection v2 32v1 Any linear combination of VI and v2 is actually just a multiple of VI For instance avl bvz avl b32v2 a 3b2v1 So Spanv1 v2 is the set of points on the line through v1 and 0 Note Exercises 19 and 20 prepare the way for ideas in Sections 14 and 17 20 Spanv1 v2 is a plane in R3 through the origin because the neither vector in this problem is a multiple of the other Every vector in the set has 0 as its second entry and so lies in the xZplane in ordinary 3space So Spanv1 v2 is the xz plane N p A Lt hTh 22 Q32 h Th39 td tr39 dt ey k enuvy 1 1 k 0 kh2 1saugmene ma 1xcorresponso a consistent system for all h and k So y is in Spanu v for all h and k 22 Construct any 3 gtlt4 matrix in echelon form that corresponds to an inconsistent system Perform sufficient row operations on the matrix to eliminate all zero entries in the rst three columns 23 21 False The alternative notation for a colunm vector is 41 3 using parentheses and commas 5 b False Plot the points to verify this Or see the statement preceding Example 3 If 2 were on 2 5 2 the 11ne through 5 and the origin then 2 would have to be a multiple of 5 which is not the case True See the line displayed just before Example 4 True See the box that discusses the matrix in 5 False The statement is often true but Spanu v is not a plane when v is a multiple of u or when u is the zero vector 95 5quot 24 a True See the beginning of the subsection Vectors in Rquot b True Use Fig 7 to draw the parallelogram determined by u v and v 0 False See the rst paragraph of the subsection Linear Combinations d True See the statement that refers to Fig 11 e True See the paragraph following the de nition of Spanv1 VF 20 CHAPTER 1 25 26 27 28 Linear Equations in Linear Algebra 21 There are only three vectors in the set 211 a2 a3 and b is not one of them b There are infmitely many vectors in W Spana1 a2 213 To determine if b is in W use the method of Exercise 13 10 4410 44 0 44 03 2103 210 21 263 406 54002 T T T T a1 a2 213 b The system for this augmented matrix is consistent so b is in W c 211 1211 0212 0213 See the discussion in the text following the de nition of Spanv1 VP 2 0 6 a1 a2 213 b 1 8 5 1 2 1 3 l 10 l 0 3 5 l 0 3 5 10 3N l 8 5 3N0 8 8 8N0 8 2 l 3 0 2 0 0 a 3 5 8 8 2 2 0 0 Yes b is a linear combination of the columns ofA that is b is in W The third column ofA is in Wbecause a3 0211 0212 1213 P a 5v1 is the output of 5 days operation of mine 1 Uquot 150 The total output is xlvl xzvz so x1 and x2 should satisfy xlv1 xzv2 2825 20 30 150 1 0 15 M Reduce the augmented matr1x N 550 500 2825 0 1 40 P Operate mine 1 for 15 days and mine 2 for 4 days This is the exact solution a The amount of heat produced when the steam plant burns x1 tons of anthracite and x2 tons of bituminous coal is 276x1 302262 million Btu b The total output produced by x1 tons of anthracite and x2 tons of bituminous coal is given by the 276 302 vector x1 3100 x2 6400 250 360 276 302 162 c M The appropriate values for x1 and x2 satisfy x1 3100 x2 6400 23610 250 360 1623 To solve row reduce the augmented matrix 276 302 162 1000 0 3900 3100 6400 23610 N 0 1000 1800 250 360 1623 0 0 0 The steam plant burned 39 tons of anthracite coal and 18 tons of bituminous coal 13 Solutions 21 29 The total mass is 2 5 2 1 10 So v 2v1 5v2 2V3 v410 That is K 5 4 4 9 1020 8 9 13 v 2 4 5 3 2 3 8 i 815 68 9 2 1 6 6 10 26 0 30 Let m be the total mass of the system By de nition 1 m v mlvlmkvk v1ivf m m m The second expression displays v as a linear combination of VI vk which shows that v is in Spanv1 vk 1 0 8 2 103 31 a The centerofmass1s 1 1 1 3 1 1 4 2 b The total mass of the new system is 9 grams The three masses added w w and W3 satisfy the equation gamiwwzmm 111121 which can be rearranged to 1112111 1113 11121 wiwiiim11121 The condition w w W3 6 and the vector equation above combine to produce a system of three equations whose augmented matrix is shown below along with a sequence of row operations 1 1 1 6 1 1 1 6 1 1 1 6 and 082 808280828 11412 0036 0012 1104 10035 10035 0804080 4010 5 0012 0012 0012 Answer Add 35 g at 0 1 add 5 g at 8 1 and add 2 g at 2 4 Extra problem Ignore the mass of the plate and distribute 6 gm at the three vertices to make the center of mass at 2 2 Answer Place 3 g at 0 1 1 g at 8 1 and 2 g at 2 4 32 See the parallelograms drawn on Fig 15 from the text Here Cl 02 03 and c4 are suitable scalars The darker parallelogram shows that b is a linear combination of VI and v2 that is 01v1 02v2 0v3 b 22 CHAPTER 1 Linear Equations in Linear Algebra The larger parallelogram shows that b is a linear combination of VI and v3 that is 04v1 0v2 03v3 b So the equation xlvl xzvz x3v3 b has at least two solutions not just one solution In fact the equation has in nitely many solutions 33 a Forj l n thejth entry of u v w is uj V w By associativity of addition in R this entry equals u VJ w which is the jth entry of u v w By de nition of equality of vectors uvwu vw For any scalar c thejth entry of Cu v is Cuj V and thejth entry of cu cv is cu CV by de nition of scalar multiplication and vector addition These entries are equal by a distributive law in R So Cu v cu cv Uquot 34 a Forj l n u iluj iluj u 0 by properties of R By vector equality u ilu ilu u 0 For scalars c and d the jth entries of Cdu and cd u are Cduj and cd uj respectively These entries in R are equal so the vectors Cdu and cdu are equal 6quot Note When an exercise in this section involves a vector equation the corresponding technology data in the data les on the web is usually presented as a set of column vectors To use MATLAB or other technology a student must rst construct an augmented matrix from these vectors The MATLAB note in the Study Guide describes how to do this The appendices in the Study Guide give corresponding information about Maple lVIathematica and the TI and HP calculators 14 SOLUTIONS Notes Key exercises are 1720 27 28 31 and 32 Exercises 29 30 33 and 34 are harder Exercise 34 anticipates the Invertible Matrix Theorem but is not used in the proof of that theorem 1 The matrixvector productAx product is not de ned because the number of columns 2 in the 3gtlt2 4 2 3 matrix 1 6 does not match the number of entries 3 in the vector 2 0 l 7 14 Solutions 23 2 The matrixvector productAx product is not de ned because the number of columns 1 in the 3gtlt1 2 5 matrix 6 does not match the number of entries 2 in the vector I 1 6 5 2 6 5 12 15 3 3 Ax 4 332 4 3 3 8 9 1and 7 7 6 14 18 4 6 5 2 6A25A 3 3 Ax 4 3 3 4A2 3A 3 1 7 6 7A26A 3 4 1 8 3 4 8 3 4 83 4 7 4 Ax 111 11 11 and 5 1 2 1 5 1 2 512 8 1 8 3 4 8A13 1 4 1 7 Ax 1 5 1 2 1 5il1A12il 8 5 On the left side of the matrix equation use the entries in the vector x as the weights in a linear combination of the columns of the matrixA 5 1 8 4 8 5 1 3 21 2 7 3 5 16 6 On the left side of the matrix equation use the entries in the vector x as the weights in a linear combination of the columns of the matrixA 7 3 1 2 1 2 5 1 9 6 12 3 2 7 47 7 The left side of the equation is a linear combination of three vectors Write the matrixA whose columns are those three vectors and create a variable vector X with three entries 4 5 7 4 5 7 x1 1 3 8 1 3 8 A and x x2 Thus the equat1onAx b 1s 7 5 0 7 5 0 4 1 7 27 4 1 2 g 4 5 7 7 6 1 3 8 xl 8 x2 7 5 0 0 9 24 CHAPTERI Linear Equations in Linear Algebra For your information The unique solution of this equation is 5 7 3 Finding the solution by hand would be timeconsuming Note The skill of writing a vector equation as a matrix equation will be important for both theory and application throughout the text See also Exercises 27 and 28 8 The left side of the equation is a linear combination of four vectors Write the matrixA whose columns are those four vectors and create a variable vector with four entries 21 4 74 75 3 4 74 75 3 z2 A and z Then the equatlon Az b 72 5 4 0 72 5 4 0 Z3 Z4 21 4 74 75 3 Z2 4 1s 72 5 4 0 z3 13 Z4 For your information One solution is 7 3 3 1 The general solution is 21 6 7523 7 12524 22 5 7 523 7 524 with 23 and Z4 free 9 The system has the same solution set as the vector equation 3 1 75 9 x1 0 x2 1 x3 4 0 and this equation has the same solution set as the matrix equation 3 1 5 xl 9 x2 0 1 4 0 x3 10 The system has the same solution set as the vector equation 8 71 4 x1 5 x2 4 l 1 73 2 and this equation has the same solution set as the matrix equation 8 71 4 5 4 x1 1 x2 1 73 2 11 To solveAx b row reduce the augmented matrix a1 a2 213 b for the corresponding linear system 124 2124 2124 2120 6 000 015 27015 27015 2010 30 073 727473900550011001100 1 14 Solutions 25 x1 0 961 0 The solution is x2 3 As a vector the solution is x x2 3 x3 1 x3 1 12 To solveAx b row reduce the augmented matrix a1 a2 213 b for the corresponding linear system 1 2 1 0 1 2 1 0 1 2 1 0 1 2 1 0 3 l2 1N055 1N05 51N0551 0 5 3 1 0 5 3 1 0 0 2 2 0 0 1 1 2 0 1 1 2 0 1 0 0 3 5 l 0 0 1 Z o 5 0 4 N 0 45 0 D 0 45 0 1 1 0 1 0 0 1 x1 35 x1 35 The solution is x2 45 As avector the solution is x x2 45 0 x3 1 x3 1 9 The vector u is in the plane spanned by the columns of A if and only if u is a linear combination of the columns of A This happens if and only if the equationAx H has a solution See the box preceding Example 3 in Section 14 To study this equation reduce the augmented matrix A u 3 5 0 1 1 4 1 l 4 1 4 2 64N 2 64N0 812N012 1 1 4 3 5 0 0 8 l2 0 0 0 The equationAx H has a solution so u is in the plane spanned by the columns ofA For your information The unique solution of Ax u is 52 32 p A A Reduce the augmented matrix A u to echelon form 5 8 7 2 l 3 0 2 1 3 0 2 3 0 2 0 1 1 3 N 0 1 1 3 N 0 1 1 3 N 0 D l 3 1 3 0 2 5 8 7 2 0 7 7 8 0 0 0 The equationAx H has no solution so u is not in the subset spanned by the columns of A 2 1 1 The augmented matrix forAx b is b1 which is row equivalent to b1 6 3 b2 0 0 b2 3bl p A UI This shows that the equationAx b is not consistent when 3b1 b2 is nonzero The set of b for which the equation is consistent is a line through the originNthe set of all points b1 b2 satisfying b2 N3b1 1 3 4 b1 16 Row reduce the augmented matrix A b A 3 2 6 b b2 5 l 8 b3 1 3 4 b1 1 3 4 b1 3 2 6 b2 N 0 7 6 b23b1 5 1 8 b3 0 14 12 b3 5b1 to o p A l p A on H N N CHAPTER 1 Linear Equations in Linear Algebra 1 3 4 b1 3 4 b1 N 0 7 6 b23b1 0 6 b23b1 0 0 0 b3 5b12b23b1 0 0 0 b12b2b3 The equationAx b is consistent if and only if b1 2b2 b3 0 The set of such b is a plane through the origin in R3 Row reduction shows that only three rows of A contain a pivot position 1 3 0 3 l 3 0 3 l 3 0 3 D 3 0 3 A l l l 1N0 2 l 4N0 2 14N0 l 4 0 4 2 8 0 4 2 8 0 0 0 0 0 0 0 2 0 3 l 0 6 3 7 0 0 0 5 0 0 0 0 Because not every row of A contains a pivot position Theorem 4 in Section 14 shows that the equation Ax b does not have a solution for each b in R4 Row reduction shows that only three rows of B contain a pivot position 1 3 2 2 l 3 2 2 l 3 2 2 D 3 2 2 B0ll 5N011 5N011 5N0 l 5 l 2 3 7 0 l l 5 0 0 0 0 0 0 0 2 8 2 l 0 2 2 3 0 0 0 7 0 0 0 0 Because not every row of B contains a pivot position Theorem 4 in Section 14 shows that the equation Bx y does not have a solution for each y in R4 The work in Exercise 17 shows that statement d in Theorem 4 is false So all four statements in Theorem 4 are false Thus not all vectors in R4 can be written as a linear combination of the columns ofA Also the columns ofA do not span R4 The work in Exercise 18 shows that statement d in Theorem 4 is false So all four statements in Theorem 4 are false Thus not all vectors in R4 can be written as a linear combination of the columns of B The columns of B certainly do not span R3 because each column of B is in R4 not R3 This question was asked to alert students to a fairly common misconception among students who are just learning about spanning Row reduce the matrix v1 v2 v3 to determine whether it has a pivot in each row 1 0 l l 0 l l 0 1 D01 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 39 0 1 1 0 1 1 0 0 1 0 0 0 The matrix v1 v2 v3 does not have a pivot in each row so the columns of the matrix do not span R4 by Theorem 4 That is v1 v2 v3 does not span R4 Note Some students may realize that row operations are not needed and thereby discover the principle covered in Exercises 31 and 32 22 23 24 N UI N CA N on 14 Solutions 27 Row reduce the matrix v1 v2 v3 to determine whether it has a pivot in each row 0 0 4 8 5 0 3 1 N 0 1 2 8 5 0 0 The matrix v1 v2 v3 has a pivot in each row so the columns of the matrix span R4 by Theorem 4 That is v1 v2 v3 spans R4 False See the paragraph following equation 3 The text calls Ax b a matrix equation True See the box before Example 3 False See the warning following Theorem 4 True See Example 4 True See parts c and a in Theorem 4 reruner True In Theorem 4 statement a is false if and only if statement d is also false 21 True This statement is in Theorem 3 However the statement is true without any quotproofquot because by de nitionAx is simply a notation for x1211 xnan where 211 an are the columns ofA Uquot True See Example 2 O True by Theorem 3 True See the box before Example 2 Saying that b is not in the set spanned by the columns of A is the same a saying that b is not a linear combination of the columns ofA 9 False See the warning that follows Theorem 4 7 5quot True In Theorem 4 statement c is false if and only if statement a is also false By de nition the matrixvector product on the left is a linear combination of the columns of the matrix in this case using weights 73 71 and 2 So cl 73 C 71 and 03 2 The equation in x1 and x2 involves the vectors u v and w and it may be viewed as u v x1 w By definition of a matrixvector product xlu xiv w The stated fact that x2 3H 7 5v 7w 0 can be rewritten as 3H 7 5v w So a solution is x1 3 x2 75 Place the vectors ql qz and q3 into the columns of a matrix say Q and place the weights x1 x2 and x3 into a vector say x Then the vector equation becomes x1 Qx v where Q ql qz q3 and x x2 x3 Note If your answer is the equationAx b you need to specify whatA and b are The matrix equation can be written as clvl czvz C3V3 c4v4 csvs v5 where cl73cz2034c471052and filmSaul itemMm 28 CHAPTER 1 0 Linear Equations in Linear Algebra 29 Start with any 3X3 matrix B in echelon form that has three pivot positions Perform a row operation a row interchange or a row replacement that creates a matrix A that is nut in echelon form Then A has the desired property The justification is given by row reducing A to B in order to display the pivot positions Since A has a pivot position in every row the columns of A span R3 by Theorem 4 30 Start with any nonzero 3X3 matrix B in echelon form that has fewer than three pivot positions Perform a row operation that creates a matrix A that is nut in echelon form Then A has the desired property Since A does not have a pivot position in every row the columns of A do not span R3 by Theorem 4 31 A 3X2 matrix has three rows and two columns With only two columns A can have at most two pivot columns and so A has at most two pivot positions which is not enough to fill all three rows By Theorem 4 the equation Ax b cannot be consistent for all b in R3 Generally if A is an mxn matrix with m gt n then A can have at most 71 pivot positions which is not enough to fill all m rows Thus the equation Ax b cannot be consistent for all b in R3 03 N A set of three vectors in cannot span R4 Reason the matrix A whose columns are these three vectors has four rows To have a pivot in each row A would have to have at least four columns one for each pivot which is not the case Since A does not have a pivot in every row its columns do not span R4 by Theorem 4 In general a set of n vectors in Rm cannot span Rm when n is less than m 03 b3 If the equation Ax b has a unique solution then the associated system of equations does not have any free variables If every variable is a basic variable then each column of A is a pivot column So the 0 0 0 G 0 0 0 39 000 reduced echelon form of A must be Note Exercises 33 and 34 are difficult in the context of this section because the focus in Section 14 is on existence of solutions not uniqueness However these exercises serve to review ideas from Section 12 and they anticipate ideas that will come later 34 If the equation Ax b has a unique solution then the associated system of equations does not have any free variables If every variable is a basic variable then each column of A is a pivot column So the 0 0 reduced echelon form of A must be 0 D 0 Now it is clear that A has a pivot position in each raw 0 0 D By Theorem 4 the columns of A span R3 35 Given Axl y1 and sz y2 you are asked to show that the equation Ax W has a solution where W y1 yz Observe that W Axl Ax2 and use Theorem 5a with X1 and x2 in place of 11 and v respectively That is W Ax1 Ax2 Ax1 x2 So the vector x x1 x2 is a solution of W Ax 36 Suppose that y and z satisfy Ay 2 Then 42 4Ay By Theorem 5b 4Ay A4y So 42 A4y which shows that 4y is a solution of Ax 42 Thus the equation AX 42 is consistent 7 2 75 8 7 2 75 8 2 75 8 37 M 75 73 4 79 0 7117 37 7237 0 37 7237 610 72 7 0 587167 17 0 0718911 0 0 77 9 215 0 ll 73 23 0 0 14 Solutions 29 2 5 8 0 429 329 or approximately 0 0 17 2 to three significant gures The original matrix does not 0 0 0 0 have a pivot in every row so its columns do not span R4 by Theorem 4 5 7 4 9 5 7 4 9 7 4 9 M 6 8 7 5 N 0 25 115 295 N 0 115 295 39 4 4 9 9 0 85 295 815 0 0 7 9 11 16 7 0 85 445 1165 0 0 MATLAB shows starred entries for numbers that are essentially zero to many decimal places So with pivots only in the rst three rows the original matrix has columns that do not span R4 by Theorem 4 12 7 11 9 5 12 7 11 9 5 M 9 4 8 7 3 N 0 54 14 14 34 6 11 7 3 9 0 152 32 32 132 4 6 10 5 12 0 113 193 2 313 12 7 11 9 5 7 11 9 5 0 54 14 14 34 0 14 14 34 N 0 0 0 0 2 N 0 0 4115 12215 0 0 285 4115 12215 0 0 0 0 The original matrix has a pivot in every row so its columns span R4 by Theorem 4 8 ll 6 7 l3 8 ll 6 7 l3 7 8 5 6 9 0 l38 l4 l8 198 I M 11 7 7 9 6 N 0 658 54 58 l9l8 3 4 l 8 7 0 658 54 438 958 8 11 6 7 13 9 11 6 7 13 0 138 14 18 198 0 14 18 198 0 0 0 0 12 0 0 0 0 000600000 The original matrix has a pivot in every row so its columns span R4 by Theorem 4 M Examine the calculations in Exercise 39 Notice that the fourth column of the original matrix sayA is not a pivot column LetA be the matrix formed by deleting column 4 of A let B be the echelon form obtained fromA and let B be the matrix obtained by deleting column 4 of B The sequence of row operations that reduces A to B also reduces A to B Since B is in echelon form it shows thatA has a pivot position in each row Therefore the columns of A span R4 It is possible to delete column 3 ofA instead of column 4 In this case the fourth column ofA becomes a pivot column of A as you can see by looking at what happens when column 3 of B is deleted For later work it is desirable to delete a nonpivot column 30 CHAPTER 1 Linear Equations in Linear Algebra Note Exercises 41 and 42 help to prepare for later work on the column space of a matrix See Section 29 or 46 The Study Guide points out that these exercises depend on the following idea not explicitly mentioned in the text when a row operation is performed on a matrixA the calculations for each new entry depend only on the other entries in the same column If a column of A is removed forming a new matrix the absence of this column has no affect on any rowoperation calculations for entries in the other columns of A The absence of a column might affect the particular choice of row operations performed for some purpose but that is not being considered here 42 M Examine the calculations in Exercise 40 The third column of the original matrix sayA is not a pivot column Let A be the matrix formed by deleting column 3 of A let B be the echelon form obtained fromA and let 3quot be the matrix obtained by deleting column 3 of B The sequence of row operations that reduces A to B also reduces A to B Since B is in echelon form it shows thatA0 has a pivot position in each row Therefore the columns of A span R It is possible to delete column 2 ofA instead of column 3 See the remark for Exercise 41 However only one column can be deleted If two or more columns were deleted fromA the resulting matrix would have fewer than four columns so it would have fewer than four pivot positions In such a case not every row could contain a pivot position and the columns of the matrix would not span R4 by Theorem 4 Notes At the end of Section 14 the Study Guide gives students a method for learning and mastering linear algebra concepts Speci c directions are given for constructing a review sheet that connects the basic de nition of span with related ideas equivalent descriptions theorems geometric interpretations special cases algorithms and typical computations I require my students to prepare such a sheet that re ects their choices of material connected with span andI make comments on their sheets to help them refine their review Later the students use these sheets when studying for exams The MATLAB box for Section 14 introduces two useful commands gauss and bgauss that allow a student to speed up row reduction while still visualizing all the steps involved The command B gauss A 1 causes MATLAB to find the leftmost nonzero entry in row 1 of matrix A and use that entry as a pivot to create zeros in the entries below using row replacement operations The result is a matrix that a student might write next to A as the first stage of row reduction since there is no need to write a new matrix after each separate row replacement I use the gauss command frequently in lectures to obtain an echelon form that provides data for solving various problems For instance if a matrix has 5 rows and if row swaps are not needed the following commands produce an echelon form ofA B gaussA1 B gaussB2 B gaussB3 B gaussB4 If an interchange is required I can insert a command such as B swap B 2 5 The command bgauss uses the leftmost nonzero entry in a row to produce zeros above that entry This command together with scale can change an echelon form into reduced echelon form The use of gauss and bg ans 5 creates an environment in which students use their computer program the same way they work a problem by hand on an exam Unless you are able to conduct your exams in a computer laboratory it may be unwise to give students too early the power to obtain reduced echelon forms with one commandithey may have difficulty performing row reduction by hand during an exam Instructors whose students use a graphic calculator in class each day do not face this problem In such a case you may wish to introduce rref earlier in the course than Chapter 4 or Section 28 which is where I finally allow students to use that command 15 SOLUTIONS Notes The geometry helps students understand Spanu v in preparation for later discussions of subspaces The parametric vector form of a solution set will be used throughout the text Figure 6 will appear again in Sections 29 and 48 15 Solutions 31 For solving homogeneous systems the text recommends working with the augmented matrix although no calculations take place in the augmented column See the Study Guide comments on Exercise 7 that illustrate two common student errors All students need the practice of Exercises 1N14 Assign all odd all even or a mixture If you do not assign Exercise 7 be sure to assign both 8 and 10 Otherwise a few students may be unable later to nd a basis for a null space or an eigenspace Exercises 2934 are important Exercises 33 and 34 help students later understand how solutions of Ax 0 encode linear dependence relations among the columns of A Exercises 3538 are more challenging Exercise 37 will help students avoid the standard mistake of forgetting that Theorem 6 applies only to a consistent equationAx b 1 Reduce the augmented matrix to echelon form and circle the pivot positions If a column of the coe icientmatrix is not a pivot column the corresponding variable is free and the system of equations has a nontrivial solution Otherwise the system has only the trivial solution 2 580 2 5 80 580 2 7100 12 900 90 4 270 012 90 0 000 The variable x3 is free so the system has a nontrivial solution 1 370 1 3 70 3 70 2 2 1 4 00 51000100 1290 0520 00 0 There is no free variable the system has only the trivial solution 3 5 7 0 5 7 0 3 N The var1able x3 1s free the system has nontr1v1al solutlons 6 7 1 0 0 15 0 An alert student will realize that row operations are unnecessary With only two equations there can be at most two basic variables One variable must be free Refer to Exercise 31 in Section 12 5 7 9 0 1 2 6 0 D 2 6 0 4 N N x3 1s a free var1able39 the system has 1 2 6 0 5 7 9 0 0 39 0 nontrivial solutions As in Exercise 3 row operations are unnecessary 1 3 1 0 1 3 1 0 1 0 5 0 0 5 0 5 4 9 20N0 3 60N03 60N0 0 0 3 60 0 3 600000 0000 5x3 0 2x3 0 The variable x3 is free x15x3 andx2N2x3 0 0 x1 SJg 5 In parametric vector form the general solution is x x2 2g x3 2 J0 0 1 32 CHAPTER 1 Linear Equations in Linear Algebra 1 3 5 0 1 3 5 0 1 3 5 0 D 0 4 0 61 4 8001 3001 300 30 3 7 9 0 0 2 6 0 0 0 0 0 0 0 0 0 4x3 0 3x3 0 The variable x3 is free x1 41x3 and x2 3x3 0 0 x1 4g 4 In parametric vector form the general solution is x x2 3 x3 3 3 3 1 713 3 7 OJNPO 9 8 0 9x3 8x40 3901 450 0 4 5039 4x35x40 The basic variables are x1 and x with x3 and x4 free Next x1 79x3 8x4 and x2 4x3 7 5x4 The general solution is x1 9x3 8x4 9g 8x4 9 8 X x2 4 5x4 4x3 5x4 x3 4 x4 5 3 x3 gt 0 0 x4 x4 0 x4 0 1 8 I 2 9 5 OJNP 0 5 7 0 69 5x3 7x4 0 390 1 2 60 0Q 2 6 039 2x3 6x40 The basic variables are x1 and x with x3 and x4 free Next x1 5x3 7x4 and x2 72263 6x4 The general solution in parametric vector form is x1 5x3 7x4 5x3 7x4 5 7 x2 2x3 6x4 2x3 6x4 2 6 x x3 x4 g x3 x3 0 1 0 x4 x4 0 x4 0 1 9 3 9 6 0 3 2 0 3 2 0 3x2 2x3 0 1 3 2 0 3 9 6 0 0 0 0 0 0 0 The solution is x1 3x2 7 2x3 with x2 and x3 free In parametric vector form 3x2 2 3x2 2x3 3 2 x x2 x2 0 x2 1 g 0 9g 0 g 0 1 10 1 3 0 4 0 3 0 4 0Q 3x2 4x40 2 6 0 8 0 0 0 0 0 0 0 0 The only basic variable is x1 so x2 x and x4 are free Note that x3 is not zero Also x1 3x2 4x4 The general solution is 15 Solutions 33 x1 3x2 4x4 3x2 4x4 3 0 4 x2 x2 x2 0 0 1 0 0 x 2x2 x4 x3 g 0 x3 0 0 l 0 x4 x4 0 0 x4 0 0 1 1 4 72 0 3 75 0 1 4 72 0 0 7 0 4 0 0 0 5 0 11 0 0 1 0 0 71 0 0 0 0 0 71 0 0 0 0 71 0 I 0 0 0 0 1 4 0 0 0 0 0 4 0 0 0 D 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7 4x 5x6 0 x6 0 4 0 The bas1c var1ables are x1 x3 and x5 The remammg var1ables are free x6 0 0 In particular 64 is free and not zero as some may assume The solution is x1 4x2 7 5x6 x3 x6 x5 4x6 with xi x4 and 66 free In parametric vector form x1 4x2 5x6 4x2 0 75x6 4 0 75 x2 x2 x2 0 0 1 0 0 x3 x6 0 0 x6 0 0 1 x x2 x4 x6 x4 x4 0 x4 0 0 1 0 x5 4x6 0 0 4x6 0 0 4 x6 x6 0 0 x6 0 0 1 T T T u v w Note The Study Guide discusses two mistakes that students often make on this type of problem 1 5 2 76 9 0 0 l 5 2 76 9 0 0 5 0 8 l 0 0 12 0 0 l 77 4 78 0 N 0 0 l 7 4 0 0 N 0 0 D 77 4 0 0 0 0 0 0 0 l 0 0 0 0 0 0 l 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 99 5x2 8x4 x5 0 7x 4x 0 4 5 0 00 The basic variables are x1 x3 and x6 the free variables are xi x4 and x5 The general solution is x1 75x2 7 8x4 7 x5 x3 7x4 7 4x5 and x6 0 In parametric vector form the solution is 34 CHAPTERI Linear Equations in Linear Algebra 7 x1 7 7 5x2 8x4 x5 7 5x2 7 7 8x4 7 x5 5 8 1 x2 x2 x2 0 0 1 0 0 x x3 2 7x4 4x5 0 7x4 4x5 2 x2 0 x4 7 x5 4 x4 x4 0 x4 0 0 1 0 x5 x5 0 x5 0 0 1 7x6 7 0 0 7 0 0 0 0 13 To write the general solution in parametric vector form pull out the constant terms that do not involve the free variable 7x17 754x3 5 74 7 5 7 4 x x2 2 7g 2 7g 2 Jg 7 px3q jgiix3 07770 71 T T P q 7 5 4 Geometrically the solution set is the line through 2 in the direction of 7 7 0 1 14 To write the general solution in parametric vector form pull out the constant terms that do not involve the free variable x1 3x4 0 3x4 0 3 x2 8 x4 8 x4 8 x x4 P 954 g 2 5x4 2 5x4 2 5 x4 x4 0 x4 0 1 T P q The solution set is the line through p in the direction of q 15 Row reduce the augmented matrix for the system 1 3 1 1 1 3 1 1 1 3 1 1 4 9 2 1 N 0 3 6 3 N 0 3 6 3 0 3 6 3 0 3 6 3 0 0 0 0 1 3 1 1 0 5 2 5x3 2 01210 2 1 2x31 0 0 0 0 0 0 0 0 0 0 Thus x1 72 5x3 x2 l 7 2x3 and x3 is free In parametric vector form x1 2 SJg 2 SJg 2 5 X x2 1 21g 1 2g 1 2 x3 x3 0 g 0 1 H 9 15 Solutions 35 72 The solution set is the line through 1 parallel to the line that is the solution set of the homogeneous 0 system in Exercise 5 Row reduce the augmented matrix for the system 137541375413754 0475 14 8 77017337017330 73 3 7377976 02 66 0000 0000 4x3 75 3x3 3 Thus x1 75 7 4x3 x2 3 3x3 and x3 is free In parametric vector form 0 0 x1 75 4x3 75 74 75 4 xx2 33x3 3 3 3x3 3 x3 x3 0 g 0 75 The solution set is the line through 3 parallel to the line that is the solution set of the homogeneous 0 system in Exercise 6 Solve x1 9x2 7 4x3 72 for the basic variable x1 72 7 9x2 4x3 with x2 and x3 free In vector form the solution is x1 72 9x2 4963 72 79x2 4x3 72 79 4 x x2 x2 0 x2 0 0 x2 lx3 0 x3 x3 0 0 x3 0 0 l The solution of x1 9x2 7 4x3 0 is x1 79x2 4x3 with x2 and x3 free In vector form x1 79x2 4x3 79x2 49 79 4 x x2 x2 x2 0 x2 1g 0 x2uojv x3 x3 0 x3 0 1 The solution set of the homogeneous equation is the plane through the origin in R3 spanned by u and v The solution set of the nonhomogeneous equation is parallel to this plane and passes through the pointp 0 0 Solve x1 7 3x2 5x3 4 for the basic variable x1 4 3x2 7 5x3 with x and 03 free In vector form the solution is x1 43x2 75x3 4 3x2 75 4 3 75 X x2 x2 0 x2 0 0 x2 1g 0 x3 g 0 0 g 0 0 1 36 CHAPTER 1 gt0 N G N p A N N The line through p and q is parallel to q p So given p Linear Equations in Linear Algebra The solution of x1 7 3x2 5x3 0 is x1 3x2 7 5x3 with x2 and x3 free In vector form x1 3x2 5x3 3x2 5x3 3 5 x x2 x2 x2 0 x2 1g 0 x2ux3v x3 x3 0 x3 0 l The solution set of the homogeneous equation is the plane through the origin in R3 spanned by u and v The solution set of the nonhomogeneous equation is parallel to this plane and passes through the pointp 0 0 The line through a parallel to b can be written as x a t b where t represents a parameter x xi 2 t 5 or xlz Z St x2 0 3 xz3t The line through a parallel to b can be written as x a tb where t represents a parameter x1 3 7 x13 7t x t or x2 4 8 x2 48t 2 3 The line through p and q is parallel to q p So given p 5 and q form 1 3 2 5 2 5 q p15 6andwritethelineasxptq p 5t 6 6 0 and q form 3 4 6 06 6 d39tthl39 6t q p 43 7 an wr1e emeasx p tq p 3 7 Note Exercises 21 and 22 prepare for Exercise 27 in Section 18 23 a True See the rst paragraph of the subsection titled Homogeneous Linear Systems 6quot False The equationAx 0 gives an implicit description of its solution set See the subsection entitled Parametric Vector Form P False The equationAx 0 always has the trivial solution The box before Example 1 uses the word nontrivial instead of trivial P False The line goes through p parallel to v See the paragraph that precedes Fig 5 5quot False The solution set could be empty The statement from Theorem 6 is true only when there exists a vector p such thatAp b 21 False A nontrivial solution ofo 0 is any nonzero x that satis es the equation See the sentence before Example 2 9 True See Example 2 and the paragraph following it N UI N CA 15 Solutions 37 0 True If the zero vector is a solution then b Ax A0 0 d True See the paragraph following Example 3 e False The statement is true only when the solution set of Ax 0 is nonempty Theorem 6 applies only to a consistent system Suppose p satis es Ax b ThenAp b Theorem 6 says that the solution set ofo b equals the set S w w p V for some vh such thatAvl1 0 There are two things to prove a every vector in S satis es Ax b b every vector that satis es Ax b is in S a Let w have the form w p vh whereAvl1 0 Then Aw Ap vh Ap Avh By Theorem 5a in section 14 b 0 b So every vector of the form p vh satis esAx b b Now let w be any solution ofo b and set V w p Then AvhAw7pAw7Ap bib0 So v1 satis es Ax 0 Thus every solution ofo b has the form w p vh Geometric argument using Theorem 6 SinceAx b is consistent its solution set is obtained by translating the solution set ofo 0 by Theorem 6 So the solution set ofo b is a single vector if and only if the solution set of Ax 0 is a single vector and that happens if and only if Ax 0 has only the trivial solution Proof using free variables If Ax b has a solution then the solution is unique if and only if there are no free variables in the corresponding system of equations that is if and only if every colunm ofA is a pivot column This happens if and only if the equationAx 0 has only the trivial solution WhenA is the 3gtlt3 zero matrix every x in R3 satis es Ax 0 So the solution set is all vectors in R3 No If the solution set ofo b contained the origin then 0 would satisfyA0 b which is not true since b is not the zero vector 21 WhenA is a 3 gtlt3 matrix with three pivot positions the equationAx 0 has no free variables and hence has no nontrivial solution 9 With three pivot positionsA has a pivot position in each of its three rows By Theorem 4 in Section 14 the equationAx b has a solution for every possible b The term quotpossiblequot in the exercise means that the only vectors considered in this case are those in R3 becauseA has three rows 21 WhenA is a 3 gtlt3 matrix with two pivot positions the equationAx 0 has two basic variables and one free variable So Ax 0 has a nontrivial solution 9 With only two pivot positions A cannot have a pivot in every row so by Theorem 4 in Section 14 the equationAx b cannot have a solution for every possible b in R3 21 WhenA is a 3 gtlt2 matrix with two pivot positions each colunm is a pivot column So the equation Ax 0 has no free variables and hence no nontrivial solution b With two pivot positions and three rows A cannot have a pivot in every row So the equationAx b cannot have a solution for every possible b in R3 by Theorem 4 in Section 14 a WhenA is a 2gtlt4 matrix with two pivot positions the equationAx 0 has two basic variables and two free variables SoAx 0 has a nontrivial solution 9 With two pivot positions and only two rows A has a pivot position in every row By Theorem 4 in Section 14 the equationAx b has a solution for every possible b in R2 L 00 CHAPTER 1 Linear Equations in Linear Algebra 2 6 33 Look at x1 7 x2 21 and notice that the second column is 3 times the rst So suitable values for 3 9 3 x1 and x2 would be 3 and 71 respectively Another pair would be 6 and 72 etc Thus x 1 satis es Ax 0 34 Inspect how the columns 211 and 212 of A are related The second column is 732 times the rst Put 3 another way 3211 2212 0 Thus 2 satis es Ax 0 Note Exercises 33 and 34 set the stage for the concept of linear dependence 35 Look forA a1 a2 213 such that 1211 1212 1213 0 That is constructA so that each row sum the sum of the entries in a row is zero 36 Look forA a1 a2 213 such that 1211 7 2212 1213 0 That is constructA so that the sum ofthe rst and third columns is twice the second column 37 Since the solution set ofo 0 contains the point 41 the vector x 41 satis es Ax 0 Write this equation as avector equation using 211 and 212 for the columns ofA 4 a1 1212 0 Then all 41211 So choose any nonzero vector for the rst column ofA and multiply that column by 7 4 1 to get the second column ofA For example set A 1 4 Finally the only way the solution set of Ax b could not be parallel to the line through 14 and the origin is for the solution set ofo b to be empty This does not contradict Theorem 6 because that theorem applies only to the case when the equationAx b has a nonempty solution set For b take any vector that is not a multiple of the columns ofA Note In the Study Guide a Checkpoint for Section 15 will help students with Exercise 37 38 No If Ax y has no solution thenA cannot have a pivot in each row SinceA is 3gtlt3 it has at most two pivot positions So the equationAx z for any z has at most two basic variables and at least one free variable Thus the solution set forAx z is either empty or has in nitely many elements 39 Ifu satis es Ax 0 thenAu 0 For any scalar 0 Theorem 5b in Section 14 shows thatAcu cAu 00 0 40 SupposeAu 0 andAv 0 Then sinceAu v Au Av by Theorem 5a in Section 14 Auv Au Av000 Now let 0 and d be scalars Using both parts of Theorem 5 Acu dv Acu Adv cAu dAv 00 d0 0 Note The MATLAB box in the Study Guide introduces the zeros command in order to augment a matrix with a column of zeros 16 Solutions 39 16 SOLUTIONS 1 Fill in the exchange table one column at a time The entries in a column describe where a sector39s output goes The decimal fractions in each column sum to 1 Distribution of Output From Goods Services Purchased by output J J input 2 7 gt Goods 8 3 gt Services Denote the total annual output in dollars of the sectors by p and p5 From the first row the total input to the Goods sector is 2 p 7 ps The Goods sector must pay for that So the equilibrium prices must satisfy income expenses PG 39ZPG 7Ps From the second row the input that is the expense of the Services sector is 8 pG 3 ps The equilibrium equation for the Services sector is income expenses pg 8PG 3Ps Move all variables to the left side and combine like terms 8 p6 7 ps 0 8pG 7 ps 0 Row reduce the augmented matrix 8 7 0 8 7 0 D 875 0 8 7 olNio 0 oiNio 0 0 The general solution is p 875 p5 with p5 free One equilibrium solution is p5 1000 and p 875 If one uses fractions instead of decimals in the calculations the general solution would be written pG 78 p5 and a natural choice of prices might be p5 80 and pG 70 Only the ratio of the prices is important pG 875 ps The economic equilibrium is unaffected by a proportional change in prices N Take some other value for p5 say 200 million dollars The other equilibrium prices are then pc 188 million pE 170 million Any constant nonnegative multiple of these prices is a set of equilibrium prices because the solution set of the system of equations consists of all multiples of one vector Changing the unit of measurement to say European euros has the same effect as multiplying all equilibrium prices by a constant The ratios of the prices remain the same no matter what currency is used 3 21 Fill in the exchange table one column at a time The entries in a column describe where a sector s output goes The decimal fractions in each column sum to 1 40 CHAPTERI Linear Equations in Linear Algebra Distribution of Output From Purchased Chemicals Fuels Machinery by output J input 2 8 4 gt Chemicals 3 1 4 gt Fuels 5 1 2 gt Machinery b Denote the total annual output in dollars of the sectors by pc pF and pm From the rst row of the table the total input to the Chemical amp Metals sector is 2 pc 8 pF 4 pm So the equillibrium prices must satisfy income expenses Pc 2Pc 8PF 4PM From the second and third rows of the table the incomeexpense requirements for the Fuels amp Power sector and the Machinery sector are respectively PF 3pc 1PF 4PM PM 5Pc 1PF 1 sz Move all variables to the left side and combine like terms Spc N8pF N4pM 0 N3pc 9pF N4pM 0 N5pc N1pF SpM 0 P M You can obtain the reduced echelon form with a matrix program Actually hand calculations are not too messy To simplify the calculations rst scale each row of the augmented matrix by 10 then continue as usual 8 8 4 0 1 1 5 0 1 1 5 0 3 9 4 0 N 3 9 4 0 N 0 6 55 0 5 1 8 0 5 1 8 0 0 6 55 0 1 1 5 0 0 1417 0 The number of decimal N 0 1 917 0 N 0 917 0 places displayed is 0 0 0 0 0 0 0 0 somewhat arbitrary The general solution is pc 1417pMpF 917pNh withpM free prM is assigned the value 100 then pc 1417 and pF 917 Note that only the ratios of the prices are determined This makes sense for if the were converted from say dollars to yen or Euros the inputs and outputs of each sector would still balance The economic equilibrium is not affected by a proportional change in prices 16 Solutions 41 4 21 Fill in the exchange table one column at a time The entries in each column must sum to 1 Distribution of Output From Agric Energy lVIanuf Transp PurChaSCd by 1 output J J J J input 65 30 30 20 gt Agric 10 10 15 10 gt Energy 25 35 15 30 gt Manuf 0 25 40 40 gt Transp Uquot Denote the total annual output of the sectors by pA pE pM and pT respectively From the rst row of the table the total input to Agriculture is 65pA 30pE 30pM 20 pT So the equilibrium prices must satisfy income expenses pA 65ng30pE30pM20pT From the second third and fourth rows of the table the equilibrium equations are pE 10pA10pE15pM10pT pM 25pA35pEl5pM30pT pT 25pE 40pM 40pT Move all variables to the left side and combine like terms 35pA 30pE 30pM 20pT 0 10pA 90pE 15pM 10pT 0 25pA 35pE 85pM 30pT 0 25pE 40pM 60pT 0 Use gauss b gauss and scale operations to reduce the augmented matrix to reduced echelon form 35 3 3 2 0 35 3 0 55 0 0 0 71 0 0 81 24 16 0 0 81 0 43 0 0 0 53 0 0 010 117 0M 0 01 117 0M 0 0 117 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Scale the first row and solve for the basic variables in terms of the free variable pp and obtain pA 203pT pE 53pT and pM 117pT The data probably justifies at most two signi cant gures so takepT 100 and round offthe other prices to pA 200pE 53 andpM 120 5 The following vectors list the numbers of atoms of boron B sulfur S hydrogen H and oxygen 0 2 0 1 0 boron 3 0 0 1 sulfur B233 H20 H3BO3 HZS 0 2 3 2 hydrogen 0 1 3 0 oxygen The coefficients in the equation xlABZS3 szHZO gt X3 H3BO3 x4AH2S satisfy 42 CHAPTER 1 9 Linear Equations in Linear Algebra 0 Move the right terms to the left side changing the sign of each entry in the third and fourth vectors and row reduce the augmented matrix of the homogeneous system 20 10020 10020 10020 100 300 10 0032 10 01 3 00 01 3 00 02 3 20N02 3 20N0032 10N0032 10 01 300 01 3 00 02 3 20 00 3 20 20 1 0 0 200 230 100 130 01 300010 20010 20 N001 230N001 230N001 230 003 20 0000000000 The general solution is x1 13 x4 x2 2x4 03 23 x4 with 64 free Take 64 3 Then x1 1 x2 6 and x3 2 The balanced equation is BZS3 6H20 gt 2H3BO3 3HZS The following vectors list the numbers of atoms of sodium Na phosphorus P oxygen 0 barium Ba and nitrogenN 3 0 0 1 sodium 1 0 2 0 phosphorus Na3PO4 4 BaNO32 6 Ba3PO42 8 NaNO3 3 oxygen 0 l 3 0 barium 0 2 0 l nitrogen The coefficients in the equation xl Na3PO4 szBaNO32 gt xyBa3POz2 xAANaNO3 satisfy 3 0 07 1 1 0 2 0 x14 962 6 963 8 x4 3 0 1 3 0 0 2 07 1 Move the right terms to the left side changing the sign of each entry in the third and fourth vectors and row reduce the augmented matrix of the homogeneous system 3 0 0 1 0 1 0 2 0 0 l 0 2 0 0 l 0 2 0 0 1 0 2 0 0 3 0 0 1 0 0 0 6 l 0 0 1 3 0 0 4 6 8 3 0N4 6 8 3 ONO 6 0 3 ONO 6 0 3 0 0 1 3 0 0 0 1 3 0 0 0 1 3 0 0 0 0 6 1 0 0 2 0 1 0 0 2 0 1 0 0 2 0 l 0 0 2 0 l 0 gt1 9 16 Solutions 43 l 0 2 0 0 l 0 2 0 0 l 0 0 l3 0 0 l 3 0 0 0 l 3 0 0 0 l 0 l2 0 N 0 0 18 3 0 N 0 0 l l6 0 N 0 0 l l6 0 0 0 6 l 0 0 0 0 0 0 0 0 0 0 0 0 0 6 l 0 0 0 0 0 0 0 0 0 0 0 The general solution is x1 13x4 x2 12x4 x3 16x4 with 96 free Take x4 6 Then x1 2 x 3 and x3 1 The balanced equation is 2Na3PO4 3BaN032 gt Ba3P042 6NaN03 The following vectors list the numbers of atoms of sodium Na hydrogen H carbon C and oxygen 0 1 0 3 0 0 sodium 1 8 5 2 0 hydrogen NaHCO3 1 H3C6H507 6 Na3C6H507 6 H20 0 C02 1 carbon 3 7 7 1 2 oxygen The order of the various atoms is not important The list here was selected by writing the elements in the order in which they rst appear in the chemical equation reading left to right x1 NaHC03 x2 H3C5H507 gt 963 Na3C6H507 x4 H20 x5 C02 The coef cients x1 x5 satisfy the vector equation 1 0 3 0 0 1 5 2 x11x2 6 x3 6 x4 0 x51 3 7 7 1 2 Move all the terms to the left side changing the sign of each entry in the third fourth and h vectors and reduce the augmented matrix 0 3000 1000 10 18 5 200 0100 130 1 6 6 0 1 omwo 010 13 0 37 7 1 20 0001 10 The general solution is x1 x5 x2 13x5 x3 13x5 x4 x5 and x5 is free Take x5 3 Then x1 x4 3 and x2 x3 1 The balanced equation is 3NaHC03 H3C6H507 gt Na3C6H507 3HZO 3COZ The following vectors list the numbers of atoms of potassium K manganese Mn oxygen 0 sulfur S and hydrogen H 1 0 0 0 2 0 potassium 1 1 0 1 0 0 manganese KMnOA 4 M150 4 H20 1 M102 2 K2504 4 H2504 4 oxygen 0 1 0 0 1 1 sulfur 0 0 2 0 0 2 hydrogen The coefficients in the chemical equation 44 CHAPTER 1 H gt0 Linear Equations in Linear Algebra xl KMnO4 xz MnSO4 X3 H20 gt satisfy the vector equation x4 MnOz xs KzSO4 Cg HzSO4 l 0 0 0 2 0 l l 0 l 0 0 x14 x2 4 Jg l x4 2 x5 4 x6 4 0 l 0 0 l l 0 0 2 0 0 2 Move the terms to the left side changing the sign of each entry in the last three vectors and reduce the augmented matrix 1 0 0 0 2 0 0 0 0 0 0 l0 0 l l 0 l 0 0 0 0 D 0 0 0 l5 0 4 4 l 2 4 4 0 N 0 0 D 0 0 l0 0 0 l 0 0 l l 0 0 0 0 D 0 25 0 0 0 2 0 0 2 0 0 0 0 0 G 5 0 The general solution is x1 x6 x2 1565 x3 x6 x4 25x6 x5 5x6 and x6 is free Take 66 2 Then x1 x3 2 and x2 3 x4 5 and x5 l The balanced equation is 2KMnO4 3MnS04 2HZO gt 5MnOz KZSO4 2H2S04 M Set up vectors that list the atoms per molecule Using the order lead Pb nitrogen N chromium Cr manganese Mn and oxygen 0 the vector equation to be solved is l 0 3 0 0 0 lead 6 0 0 0 0 l nitrogen x10 x21Jg 0 x4 2 x5 0 x6 0 chromium 0 2 0 0 l 0 manganese 0 8 4 3 2 1 oxygen The general solution is x1 l6x5 x2 2245x5 x3 118x5 x4 ll45mg x5 4445x6 and x6 is free Take 66 90 Then x1 15 x2 44 x3 5 x4 22 and x5 88 The balanced equation is lSPbN6 44CranOg gt 5Pb304 22Cr203 88Mn02 90NO M Set up vectors that list the atoms per molecule Using the order manganese Mn sulfur S arsenic As chromium Cr oxygen 0 and hydrogen H the vector equation to be solved is l 0 0 l 0 0 0 manganese l 0 l 0 3 0 sulfur 0 2 0 0 l 0 0 arsenic x1 x2 x3 x4 x5 x6 x7 0 10 0 0 0 l 0 chrom1um 0 35 4 4 0 l2 1 oxygen 0 0 2 l 3 0 2 hydrogen In rational format the general solution is x1 l6327x7 x2 l3327x7 x3 374327x7 x4 l6327x7 x5 26327x7 x6 l30327x7 and x7 is free Take x7 327 to make the other variables whole numbers The balanced equation is l6MnS l3As2Cr10035 374H2SO4 l6HlVInO4 26AsH3 l30CrS3012 327HZO H N l 6 Solutions 45 Note that some students may use decimal calculation and simply quotround offquot the fractions that relate x1 66 to x7 The equations they construct may balance most of the elements but miss an atom or two Here is a solution submitted by two of my students 5MnS 4As2Cr10035 115H2SO4 gt 5HlInO4 8AsH3 40Cr83012 100HZO Everything balances except the hydrogen The right side is short 8 hydrogen atoms Perhaps the students thought that the 4H2 hydrogen gas escaped Node Flow in A x1 g B x2 C 80 Total ow 80 Rearrange the equations x1 x2 x1 x2 Reduce the augmented matrix 1 0 l 0 20 0 1 1 1 0 1 1 0 0 80 0 0 0 1 60 Write the equations for each node Flow out A 20 x 20 3 1 B X4 x4 x1 x2 80 C X2 X4 20 20 x4 0 80 x4 60 Do 1020 0 1060 N000 60 00 00 0 For this type of problem the best description of the general solution uses the style of Section 12 rather than parametric vector form x1 20 x3 x2 60x3 x3 is free X460 Since x1 cannot be negative the largest value of x3 is 20 Write the equations for each intersection Intersection Flow in A x1 B 200 C x2 g D x4 x5 Total ow 200 Flow out g x4 40 x1 x2 x5 100 60 200 46 CHAPTERI Linear Equations in Linear Algebra Rearrange the equations x1 g x4 40 x1 x2 200 x2 9 x5 100 x4 x5 60 Reduce the augmented matrix 1 0 1 1 0 40 D 0 1 0 1 100 1 1 0 0 0 200 0 1 0 1 100 0110 1100000 160 0001160000000 The general solution written in the style of Section 12 is xl100g x5 xl40x3 xz100 x3x5 xzl60 x3 g is free b When x4 0 X5 must be 60 and x3 is free x4 60 x5 x4 0 x5 is free x5 60 c The minimum value of x1 is 40 carsminute because x3 cannot be negative 13 Write the equations for each intersection Intersection Flow in Flow out A x2 30 x1 80 B x3 x5 x2 x4 C x6 100 x5 40 D x4 40 x6 90 E x1 60 x3 20 Total ow 230 230 Rearrange the equations x1 x2 50 x2 g x4 x5 0 x5 x6 60 x4 x6 50 x1 9g 40 Reduce the augmented matrix 1 1 0 0 0 0 50 1 1 0 0 0 0 50 0 1 1 1 1 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 60 A 0 0 0 1 0 1 50 0 0 0 1 0 1 50 0 0 0 1 1 60 1 0 1 0 0 0 40 0 0 0 0 0 0 0 a The general solution is 0 0 0 40 0 0 0 10 0 1 50 0 1 60 0 0 0 0 ampm x29g10 x3isfree x4x650 x6isfree 16 Solutions b To nd minimum ows note that since x1 cannot be negative x3 3 40 This implies that x 3 50 Also since 66 cannot be negative 64 3 50 and x5 3 60 The minimum ows are x2 50 x3 40 x4 50 x5 60 when x1 0 andxg 0 14 Write the equations for each intersection Intersection Flow in Flow out A x1 x2 100 B x2 50 x3 C x3 x4 120 D x4 150 x5 E x5 x6 80 F x6 100 x1 Rearrange the equations x1 x2 x2 x3 x3 x4 x4 x5 x5 x6 99 x6 Reduce the augmented matrix 1 1 0 0 0 0 100 l 0 1 1 0 0 0 50 0 0 0 1 1 0 0 120 0 0 0 0 1 1 0 150 MW 0 0 0 0 0 1 1 80 0 1 0 0 0 0 l 100 0 OOOOD I 120 150 100 50 120 150 100 0000 100 120 150 47 48 CHAPTERl Linear Equations in Linear Algebra 1 0 0 0 0 1 100 x1100x6 0 1 0 0 0 1 0 x2 x6 0 0 1 0 0 l 5 x3 50 x6 N i N The general solution is 0 0 0 1 0 l 70 x4 70 x6 0 0 0 0 1 1 80 x5 80 x6 0 0 0 0 0 0 0 x6 is free Since x4 cannot be negative the minimum value of 66 is 70 Note The MATLAB box in the Study Guide discusses rational calculations needed for balancing the chemical equations in Exercises 9 and 10 As usual the appendices cover this material for lVIaple Mathematica and the TI and HP graphic calculators 17 SOLUTIONS Note Key exercises are 9720 and 23730 Exercise 30 states a result that could be a theorem in the text There is a danger however that students will memorize the result without understanding the proof and then later mix up the words row and column Exercises 37 and 38 anticipate the discussion in Section 19 of onetoone transformations Exercise 44 is fairly difficult for my students 1 Use an augmented matrix to study the solution set of xlu xiv xjw 0 where u v and w are the 5 7 9 0 7 9 0 three given vectors Since 0 2 4 0 N 0 4 0 there are no free variables So the 0 6 8 0 0 0 0 homogeneous equation has only the trivial solution The vectors are linearly independent 2 Use an augmented matrix to study the solution set of xlu xiv X3W 0 where u v and w are the 0 0 3 0 Q 8 1 0 three given vectors Since 0 5 4 0 N 0 4 0 there are no free variables So the 2 8 1 0 0 0 0 homogeneous equation has only the trivial solution The vectors are linearly independent 43 Use the method of Example 3 or the box following the example By comparing entries of the vectors one sees that the second vector is 73 times the first vector Thus the two vectors are linearly dependent A 1 2 From the first entries in the vectors it seems that the second vector of the pair 4 8 may be 2 times the first vector But there is a sign problem with the second entries So neither of the vectors is a multiple of the other The vectors are linearly independent UI Use the method of Example 2 Row reduce the augmented matrix forAx 0 0 8 5 0 l 3 2 0 l 3 2 0 l 3 2 3 7 4 0 3 7 4 0 0 2 2 0 0 2 2 15 40N 15 40 0 2 20N0 0 0 l 3 2 0 0 8 5 0 0 8 5 0 0 0 3 3 0 0 0 I OGDNN oooo Ch gt1 9 17 Solutions 49 There are no free variables The equationAx 0 has only the trivial solution and so the columns of A are linearly independent Use the method of Example 2 Row reduce the augmented matrix forAx 0 4 300103010301030 030 0 140 0 1400 1400 140040 1030N 4 300N0 3120N0 000N0 0 0 5460546004 9000700000 There are no free variables linearly independent The equationAx 0 has only the trivial solution and so the columns of A are Study the equationAx 0 Some people may start with the method of Example 2 1 4 3 0 0 1 4 3 0 0 4 3 0 0 2 7 5 1 0 N 0 1 1 1 0 N 0 D 1 1 0 4 5 7 5 0 0 11 5 5 0 0 0 6 0 But this is a waste of time There are only 3 rows so there are at most three pivot positions Hence at least one of the four variables must be free So the equationAx 0 has a nontrivial solution and the columns of A are linearly dependent Same situation as with Exercise 7 The unnecessary row operations are 1 33 201 33 20 33 20 3 7 1 2 00 2 8 4008 40 01 43001 4300000 Again because there are at most three pivot positions yet there are four variables the equationAx 0 has a nontrivial solution and the columns of A are linearly dependent F The vector v3 is in Spanv1 v2 if and only if the equation xlv1 xzvz v3 has a solution To nd out row reduce v1 v2 v3 considered as an augmented matrix 1 3 5 3 5 3 9 7 N 0 0 2 6 h 0 0 h 10 At this point the equation 0 8 shows that the original vector equation has no solution So v3 is in Spanv1 v2 for no value of h b For v1 v2 v3 to be linearly independent the equation xlvl xzvz x3v3 0 must have only the trivial solution Row reduce the augmented matrix v1 v2 v3 0 1 3 501 3 5 0 350 3 9 700 0 8 00 00 2 6h0 0011 100 0000 For every value of h x is a free variable and so the homogeneous equation has a nontrivial solution Thus v1 v2 v3 is a linearly dependent set for all h 50 CHAPTERl Linear Equations in Linear Algebra 10 a The vector v3 is in Spanv1 v2 if and only if the equation xlv1 xzvz v3 has a solution To nd out row reduce v1 v2 v3 considered as an augmented matrix 1 2 2 D 2 2 5 10 9 N 0 0 3 6 h 0 0 h6 At this point the equation 0 1 shows that the original vector equation has no solution So v3 is in Spanv1 v2 for no value of h P For v1 v2 v3 to be linearly independent the equation xlvl xzvz x3v3 0 must have only the trivial solution Row reduce the augmented matrix v1 v2 v3 0 1 2 2 0 1 2 2 0 2 2 0 510 9 0N0 010N0 0 0 3 6 h 0 0 0 h 6 0 0 0 0 0 For every value of 11 x2 is a free variable and so the homogeneous equation has a nontrivial solution Thus vb v2 v3 is a linearly dependent set for all h 11 To study the linear dependence of three vectors say v1 v2 v3 row reduce the augmented matrix V1 V2 V3 0 1 3 1 0 1 3 1 0 D 3 1 0 1 5 50N0 2 4 0N0 4 0 4 7 h 0 0 5 h 4 0 0 0 h 6 0 The equation xlvl xzvz x3v3 0 has a nontrivial solution if and only if h 7 6 0 which corresponds to x3 being a free variable Thus the vectors are linearly dependent if and only if h 6 N To study the linear dependence of three vectors say v1 v2 v3 row reduce the augmented matrix v1 v2 v3 0 2 6 8 0 6 8 0 4 7 h 0 N 0 h 16 0 l 3 4 0 0 0 0 0 The equation xlv1 xiv2 205v3 0 has a free variable and hence a nontrivial solution no matter what the value of 11 So the vectors are linearly dependent for all values of h 543 To study the linear dependence of three vectors say v1 v2 v3 row reduce the augmented matrix V1 V2 V3 0 1 2 3 0 D 2 3 0 5 9 h 0 N 0 G h 15 0 3 6 9 0 0 0 0 0 The equation xlv1 xzvz x3v3 0 has a free variable and hence a nontrivial solution no matter what the value of 11 So the vectors are linearly dependent for all values of h 14 p A Ch gt1 9 gt0 21 22 17 Solutions 51 To study the linear dependence of three vectors say v1 v2 v3 row reduce the augmented matrix v1 v2 v3 0 1 5101 510 510 171002200 20 3 8 h 0 0 7 h3 0 0 0 h10 0 The equation xlvl xzvz x3v3 0 has a nontrivial solution if and only if h 10 0 which corresponds to x3 being a free variable Thus the vectors are linearly dependent if and only 39 h 710 The set is linearly dependent by Theorem 8 because there are four vectors in the set but only two entries in each vector The set is linearly dependent because the second vector is 32 times the first vector The set is linearly dependent by Theorem 9 because the list of vectors contains a zero vector The set is linearly dependent by Theorem 8 because there are four vectors in the set but only two entries in each vector The set is linearly independent because neither vector is a multiple of the other vector Two of the entries in the first vector are 7 4 times the corresponding entry in the second vector But this multiple does not work for the third entries The set is linearly dependent by Theorem 9 because the list of vectors contains a zero vector False A homogeneous system always has the trivial solution See the box before Example 2 a b False See the warning after Theorem 7 0 True See Fig 3 after Theorem 8 d True See the remark following Example 4 a True See Fig 1 1 2 b False For instance the set consisting of 2 and 74 is linearly dependent See the warning after 3 6 Theorem 8 0 True See the remark following Example 4 d False See Example 3a I gtxlt gtxlt I 0 I l 0 l 0 0 0 l 0 0 0 l 24 25 and 0 0 0 0 0 0 0 0 0 0 0 0 l 0 0 0 0 v to DJ p A 43 N 43 DJ 43 A 43 Ch DJ on CHAPTER 1 Linear Equations in Linear Algebra I 0 I 0 0 I The columns must lmearly independent by Theorem 7 because the first column 1s not 0 0 0 zero the second column is not a multiple of the first and the third colunm is not a linear combination of the preceding two columns because a3 is not in Spana1 a2 All ve colunms of the 7gtlt5 matrixA must be pivot columns Otherwise the equationAx 0 would have a free variable in which case the colunms of A would be linearly dependent If the colunms of a 5gtlt7 matrixA span R5 thenA has a pivot in each row by Theorem 4 Since each pivot position is in a different column A has ve pivot columns A any 3gtlt2 matrix with two nonzero colunms such that neither column is a multiple of the other In this case the colunms are linearly independent and so the equationAx 0 has only the trivial solutlon B any 3gtlt2 matrix with one colunm a multiple of the other 21 b The colunms of A are linearly independent if and only if the equationAx 0 has only the trivial solution This happens if and only if Ax 0 has no free variables which in turn happens if and only if every variable is a basic variable that is if and only if every colunm ofA is a pivot column Think of A a1 a2 213 The text points out that 213 a1 a2 Rewrite this as 211 a2 7 a3 0 As a matrix equationAx 0 for x l l 71 Think ofA a1 a2 213 The text points out that a1 2212 a3 Rewrite this as 211 2212 7 a3 0 As a matrix equation Ax 0 for x l 2 fl True by Theorem 7 The Study Guide adds another justification True by Theorem 9 False The vector v1 could be the zero vector False Counterexample Take v1 v2 and v4 all to be multiples of one vector Take v3 to be not a multiple of that vector For example 1 2 l 4 l 2 0 4 V115V2 2 9V3 0 9V4 4 l 2 0 4 True A linear dependence relation among v1 v2 v3 may be extended to a linear dependence relation among v1 v2 v3 v4 by placing a zero weight on v4 True If the equation xlvl xzvz x3v3 0 had a nontrivial solution with at least one of x1 x2 x3 nonzero then so would the equation xlvl xzvz x3v3 0 v4 0 But that cannot happen because v1 v2 v3 v4 is linearly independent So v1 v2 v3 must be linearly independent This problem can also be solved using Exercise 37 if you know that the statement there is true 17 Solutions 53 39 If for all h the equationAx b has at most one solution then take b 0 and conclude that the equation Ax 0 has at most one solution Then the trivial solution is the only solution and so the columns of A are linearly independent variables Ifthere is a solution it must be umque 8 3 0 7 2 8 3 0 7 2 9 5 ll 7 0 58 5 258 l94 M A N 6 2 2 4 4 0 14 2 54 52 5 l 7 0 10 0 78 7 358 354 8 3 0 7 2 Q 3 0 7 2 0 58 5 258 194 0 5 258 194 0 0 0 0 225 0 0 0 0 0 0 0 0 775 0 0 0 0 0 8 9 The p1vot columns of A are l 2 and 5 Use them to form B 6 5 8 0 9 5 Other 11ker cho1ces use colunms 3 or 4 ofA 1nstead of 2 2 5 7 10 An an matrix with n pivot colunms has a pivot in each column So the equationAx b has no free 8 7 2 11 7 6 4 4 I 5 0 10 Actually any set of three columns of A that includes column 5 will work for B but the concepts needed to prove that are not available now Column 5 is not in the twodimensional subspace spanned by the rst four columns M 1210 6 3 710 10 6 3 7 7 6 4 7 9 5 012 214 5912 9 9 9 5 5 1m 0 0 0 892 4 3 1 6 8 9 0 0 0 0 0 8 7 5 9 11 8 0 0 0 0 0 12 10 7 6 The pivot columns ofA are 1 2 4 and 6 Use them to form B 9 9 4 3 8 7 10 656 0 Other likely choices might use column 3 ofA instead of 2 andor use column 5 instead of 4 54 CHAPTER 1 Linear Equations in Linear Algebra 43 M Make v any one of the columns of A that is not in B and row reduce the augmented matrix B v The calculations will show that the equation Bx v is consistent which means that v is a linear combination of the columns of B Thus each column ofA that is not a column of B is in the set spanned by the colunms of B 44 M Calculations made as for Exercise 43 will show that each colunm ofA that is not a colunm of B is in the set spanned by the columns of B Reason The original matrixA has only four pivot colunms If one or more colunms of A are removed the resulting matrix will have at most four pivot columns Use exactly the same row operations on the new matrix that were used to reduceA to echelon form If v is a colunm ofA that is not in B then row reduction of the augmented matrix B v will display at most four pivot columns Since B itself was constructed to have four pivot colunms adjoining v cannot produce a h pivot colunm Thus the first four colunms of B v are the pivot colunms This implies that the equation Bx v has a solution Note At the end of Section 17 the Study Guide has another note to students about Mastering Linear Algebra Concepts The note describes how to organize a review sheet that will help students form a mental image of linear independence The note also lists typical misuses of terminology in which an adjective is applied to an inappropriate noun This is a major problem for my students I require my students to prepare a review sheet as described in the Study Guide and I try to make helpful comments on their sheets I am convinced through personal observation and student surveys that the students who prepare many of these review sheets consistently perform better than other students Hopefully these students will remember important concepts for some time beyond the nal exam 18 SOLUTIONS Notes The key exercises are 17720 25 and 31 Exercise 20 is worth assigning even if you normally assign only odd exercises Exercise 25 and 27 can be used to make a few comments about computer graphics even if you do not plan to cover Section 26 For Exercise 31 the Study Guide encourages students not to look at the proof before trying hard to construct it Then the Guide explains how to create the proof Exercises 19 and 20 provide a natural segue into Section 19 I arrange to discuss the homework on these exercises when I am ready to begin Section 19 The de nition of the standard matrix in Section 19 follows naturally from the homework and so I ve covered the rst page of Section 19 before students realize we are working on new material The text does not provide much practice determining whether a transformation is linear because the time needed to develop this skill would have to be taken away from some other topic If you want your students to be able to do this you may need to supplement Exercises 29 30 32 and 33 If you skip the concepts of onetoone and onto in Section 19 you can use the result of Exercise 31 to show that the coordinate mapping from a vector space onto Rquot in Section 44 preserves linear independence and dependence of sets of vectors See Example 6 in Section 44 Tuareg ll lfltvgtli illZHEZl 5 0 0 1 5 5 0 1 5a 2 TuAu 0 5 0 0 0 Tv 0 5 0 b 5b 0 0 5 4 2 0 5 0 5c 1 0 l 0090 oon 1 0 2 1 1 0 2 1 1 0 2 3Ab 2 16 N012 5N012 3 2 5 3 0 2 1 0 0 0 5 1 1 0 2 1 1 0 0 3 3 N 0 l 2 5 N 0 l 0 l x 1 unique solution 0 0 1 2 0 0 1 2 2 1 3 2 6 1 3 2 6 1 3 2 4A b0 1 4 7N0 1 4 7N0 1 4 3 5 9 9 0 4 15 27 0 0 l 1 3 0 4 0 0 5 5 N 0 l 0 3 N 0 l 0 3 x 3 unique solution 0 0 1 1 0 0 1 1 1 1 7 2 1 5 7 2 0 3 3 5 A b N N 3 7 5 2 0 1 2 1 0 D 2 1 3 Note that a solution is not 1 To avoid this common error write the equations 3 3 3 3 xl x3 and solve for the bas1c variables x2 l 2 2 l x3 is free x1 3 3Jg 3 3 General solution x x2 l 2Jg x3 g 0 l 03 0 and x l 1 2 1 1 1 2 1 1 1 2 1 l 3 4 5 9 0 2 2 6 0 1 1 3 6 A b N N 0 1 1 3 0 1 l 3 0 0 0 0 3 5 4 6 0 1 l 3 0 0 0 0 7 3 3 7 xi x2 3 g a 3 x3 is free x1 7 3Jg 7 3 7 General solution x x2 3 x3 3 963 1 one choice 3 x3 x3 0 l 0 18 39 OOWQ Solutions 1 x3 2 For a particular solution one might choose 55 56 CHAPTERI Linear Equations in Linear Algebra 7 a 5 the domain ofT is R5 because a 6gtlt5 matrix has 5 columns and forAx to be de ned x must be in R5 b 6 the codomain of T is R6 becauseAx is a linear combination of the columns of A and each column ofA is in R6 8 A must have 5 rows and 4 columns For the domain of T to be R4A must have four columns so thatAx is de ned for x in R4 For the codomain of T to be R5 the columns of A must have ve entries in which caseA must have ve rows becauseAx is a linear combination of the columns of A 1 4 7 5 0 1 4 7 5 0 1 4 7 5 0 9 SolveAx0 0 1 4 3 0 N 0 1 4 0 N 0 1 4 3 0 2 6 6 4 0 0 2 8 6 0 0 0 0 0 0 9 7 0 970 9g7x40x1x3x4 xz4g 3x4 N0 430 4g3x40 glsfree 0 0 0 0 0 0 0 x41sfree x1 9Jg 7x4 9 7 x2 4x3 3x4 4 3 x g x4 x3 x3 1 0 x4 x4 0 1 1 3 9 2 0 1 3 9 2 0 1 3 9 2 0 1 0 3 4 0 0 3 6 6 0 0 1 2 3 0 10 SolveAx 0 N 0 1 2 3 0 0 1 2 3 0 0 3 6 6 0 2 3 0 5 0 0 9 18 9 0 0 9 18 9 0 1 3 9 2 0 1 3 9 0 0 D 0 3 0 0 0 1 2 3 0 0 1 2 0 0 0 2 0 0 0 0 0 3 0 0 0 0 1 0 0 0 0 0 0 0 0 18 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 0 xl x3 x3 xz 2x3 2g 2 2 0 x 2 x3 1s free x3 0 x4 0 0 11 Is the system represented by A b consistent Yes as the following calculation shows 1 4 7 5 1 1 4 7 5 1 4 7 5 1 0 1 4 3 1 N 0 1 4 3 1 N 0 D 4 3 1 2 6 6 4 0 0 2 8 6 2 0 0 0 0 0 The system is consistent so b is in the range of the transformation X I gt AX 18 39 Solutions 57 12 Is the system represented by A b consistent 1392 11392 11392 1 103 430 3 6 640123 1 012 3 1N0 12 3 1N0 3 6 6 4 23054091892091892 139 2 11392 1 012 3 10123 1 N00031N00031 000 1811000017 The system is inconsistent so b is not in the range of the transformation X I gt AX A re ection through the origin A contraction by the factor 5 The transformation in Exercise 13 may also be described as a rotation of n radians about the origin or a rotation of in radians about the origin 15 A projection onto the xzaXis A re ection through the line x2 x1 2 6 1 2 17 T3u 3Tu 313 T2v 2Tv 2 3 6 and T3u 2v 3Tu 2Tv g 2 3 v 00 gt0 N G 21 22 CHAPTER 1 Linear Equations in Linear Algebra Draw a line through w parallel to v and draw a line through w parallel to u See the left part of the gure below From this estimate that w u 2v Since Tis linear Tw Tu 2Tv Locate Tu and 2Tv as in the right part of the gure and form the associated parallelogram to locate T w x 2Tv X2 2 w u Tv TW 2v v X1 X1 Tu All we know are the images of el and e2 and the fact that T is linear The key idea is to write 5 l 0 x 3 0 3 I 5e1 3e2 Then from the linearity ofT write T T5 3 5T 3T 5 3 52 3 1 13 X 91 92 90 92 YI y2 5 4 7 1 0 To nd the image of x1 observe that x x1 x1 x2 qu xze2 Then x2 x2 0 1 2 1 2x1 x2 TX Txlel Jr16292 x1Te1 x2Tez C15 x2 6 5x1 6x 2 Use the basic de nition of Ax to constructA Write 2 7 2 7 Txx1vlxzv2v1 v21 x A 2 5 3 5 3 a True Functions from Rquot to R are de ned before Fig 2 A linear transformation is a function with certain properties 6quot False The domain is R5 See the paragraph before Example 1 0 False The range is the set of all linear combinations of the columns of A See the paragraph before Example 1 Q False See the paragraph after the de nition of a linear transformation True See the paragraph following the box that contains equation 4 D a True See the paragraph following the de nition of a linear transformation False If A is an an matrix the codomain is R39quot See the paragraph before Example 1 Uquot 0 False The question is an existence question See the remark about Example 1d following the solution of Example 1 Q True See the discussion following the de nition of a linear transformation 05 True See the paragraph following equation 5 29 18 Solutions 59 X2 u V I1 7 X1 Tv Tu Tu v Given any x in Rquot there are constants cl cp such that x clvl cpvp because v1 vp span Rquot Then from property 5 of a linear transformation Tx clTv1 cpTvp 010 CFO 0 Any point x on the line through p in the direction of v satisfies the parametric equation x p tv for some value of I By linearity the image T x satisfies the parametric equation TXTIDIVTIDITV If Tv 0 then T x T p for all values of t and the image of the original line is just a single point Otherwise is the parametric equation of a line through T p in the direction of T v Any point x on the plane P satis es the parametric equation x su tv for some values of s and I By linearity the image T x satis es the parametric equation Tx sTu tTv 3 tin R The set of images is just SpanTu T v If T u and T v are linearly independent SpanTu T v is a plane through T u T v and 0 If T u and T v are linearly dependent and not both zero then SpanTu Tv is a line through 0 IfTu Tv 0 then SpanTu Tv is 0 a From Fig 7 in the exercises for Section 15 the line through T p and T q is in the direction of q p and so the equation ofthe line is x p tq7p ptq ftp 1 itp tq b Consider x 17 tp tq for t such that 0 E t 1 Then by linearity of T T01T1IPIQ1ITPHTQ Oitil If T p and T q are distinct then is the equation for the line segment between T p and T q as shown in part a Otherwise the set of images is just the single point T p because 1 ITP t ITQ1 tTID t ITP TID Consider a point x in the parallelogram determined by u and v say x au bv for 0 E a E 1 0 E b E 1 By linearity of T the image of x is TxTaubvaTubTvfor0a10b1 This image point lies in the parallelogram determined by T u and T v Special degenerate cases arise when T u and T v are linearly dependent If one of the images is not zero then the parallelogram is actually the line segment from 0 to T u T v If both T u and T v are zero then the parallelogram is just 0 Another possibility is that even 11 and v are linearly dependent in which case the original parallelogram is degenerate either a line segment or the zero vector In this case the set of images must be degenerate too a When b 0fx m In this case for all xy in R and all scalars c and d fcx dy mcx dy max mdy come deny cfx W This shows that f is linear o o 43 6 DJ 43 N 43 DJ 43 A 43 UI 43 Ch 1 CHAPTER 1 Linear Equations in Linear Algebra b When f x mx b with b nonzero f0 m0 b b 0 This shows that f is not linear because every linear transformation maps the zero vector in its domain into the zero vector in the codomain In this case both zero vectors are just the number 0 Another argument for instance would be to calculate f 226 m2x b and 2f x 2m 2b If b is nonzero then f 226 is not equal to 2f x and so f is not a linear transformation 0 In calculus f is called a linear function because the graph of f is a line Let Tx Ax b for x in Rquot If h is not zero T0 A0 b b 0 Actually Tfails both properties of a linear transformation For instance T 2x A2x b 2Ax b which is not the same as 2Tx 2Ax b 2Ax 2b Also TxyAxy bAx Ay b which is not the same as TxTyAxbAyb The Study Guide has a more detailed discussion of the proof Suppose that v1 v2 v3 is linearly dependent Then there exist scalars cl cl 03 not all zero such that clvl czvz C3V3 0 Then Tclv1 czv2 C3V3 T0 0 Since T is linear 01TV1 02TV2 03TV3 0 Since not all the weights are zero T v1 T v T v3 is a linearly dependent set Take any vector x1 x2 with x2 0 and use a negative scalar For instance T 0 l 72 3 but T7l0 1 T0 fl 2 3 ilT0 l One possibility is to show that T does not map the zero vector into the zero vector something that every linear transformation does do T 0 0 0 4 0 Suppose that u v is a linearly independent set in R and yet T u and T v are linearly dependent Then there exist weights cl CZ not both zero such that clTu czTv 0 Because T is linear T clu 02v 0 That is the vector x clu 02v satis es T x 0 Furthermore x cannot be the zero vector since that would mean that a nontrivial linear combination of u and v is zero which is impossible because u and v are linearly independent Thus the equation T x 0 has a nontrivial solution Take u and v in R3 and let 0 and dbe scalars Then cu dv cul dvl cuz dvz cug dVg The transformation T is linear because Tcu dv cul dvl cuz dvz 7 CT 3 dv3 cul dvl cuz dvz cug dV3 C711 C712 cu3dV1 de 97 Cula 2 3 dV1 V2 V3 cTu dTv Take u and v in R3 and let 0 and dbe scalars Then cu dv cu1 dvl cuz dvz 0143 dVg The transformation T is linear because Tcu dv cul dvl 0 Mg dV3 cm 0 cug dvl 0 dV3 Cul 0 M3 dv1 0 v3 cTu dTv 18 Solutions 61 4 2 5 5 0 D 0 0 72 0 x172x4 72 9 7 8 0 0 0 0 92 0 92 92 37 M N D x2 x xx4 6 4 5 3 0 0 0 CD 0 0 x3 0 0 7 5 3 8 4 07 70 0 0 0 0 x4 is free 1 9 4 9 4 0 D 0 0 34 0 x1 34x4 34 8 7 6 0 0 0 54 0 54 54 38 M N D x2 x xx4 11 16 9 0 0 0 74 0 x374x4 74 7 9 7 4 5 07 70 0 0 0 0 x4 is free 1 4 2 5 5 7 D 0 0 72 4 9 7 8 0 5 0 0 92 7 39 M N yes his in the range of the transformation 6 4 5 3 9 0 0 a 0 1 5 3 8 4 7 0 0 0 0 0 because the augmented matrix shows a consistent system In fact x1 4 7 2x4 4 x2 792x4 7 the general solutlon 1s 1 when x4 0 a solutlon 1s x 1 x4 is free 0 9 4 9 4 7 0 0 34 54 8 7 6 7 0 D 0 54 114 40 M N yes b 1s 1n the range of the 7 11 16 9 13 0 0 CD 74 134 9 7 4 5 5 0 0 0 0 0 transformation because the augmented matrix shows a consistent system In fact x1 54 34x4 2 x2 114 54x4 the general solutlon 1s when x4 1 a solutlon 1s x x3 13474x4 x4 is free 1 Notes At the end of Section 18 the Study Guide provides a list of equations figures examples and connections with concepts that will strengthen a student s understanding of linear transformations I encourage my students to continue the construction of review sheets similar to those for span and linear independence but I refrain from collecting these sheets At some point the students have to assume the responsibility for mastering this material If your students are using MATLAB or another matrix program you might insert the de nition of matrix multiplication after this section and then assign a project that uses random matrices to explore properties of matrix multiplication See Exercises 3L36 in Section 21 Meanwhile in class you can continue with your plans for nishing Chapter 1 When you get to Section 21 you won t have much to do The Study Guide s MATLAB note for Section 21 contains the matrix notation students will need for a project on matrix multiplication The appendices in the Study Guide have the corresponding material for Mathematica Maple and the T83 86 89 and HP48G graphic calculators 62 CHAPTER 1 Linear Equations in Linear Algebra 19 SOLUTIONS Notes This section is optional if you plan to treat linear transformations only lightly but many instructors will want to cover at least Theorem 10 and a few geometric examples Exercises 15 and 16 illustrate a fast way to solve Exercises 17 22 without explicitly computing the images of the standard basis The purpose of introducing onetoone and onto is to prepare for the term isomorphism in Section 44 and to acquaint math majors with these terms Mastery of these concepts would require a substantial digression and some instructors prefer to omit these topics and Exercises 25 40 In this case you can use the result of Exercise 31 in Section 18 to show that the coordinate mapping from a vector space onto R in Section 44 preserves linear independence and dependence of sets of vectors See Example 6 in Section 44 The notions of onetoone and onto appear in the Invertible Matrix Theorem Section 23 but can be omitted there if desired Exercises 25 28 and 31 36 offer fairly easy writing practice Exercises 31 32 and 35 provide important links to earlier material 3 5 1 2 1ATe1 Te2 3 0 1 1 4 5 2ATei T 32 Te3l3 7 4 0 1 3 Te1 e2 Te2 e1A e2 e11 0 1M 1JE 1JE 1N5 4 T l 2 3T 2 3A 6 AAii e IA2i AAE 1MB 5 T 2 1 T A 1 0 ei ei 62 2 3 CD 23 2 1 6 T T 3 3 A 1 3 CO eh ez ez 31 1 3 0 1 7 Follow what happens to e1 and e2 Since e1 is on the unit circle in the plane it rotates through 37r4 radians into a point on the unit circle that lies in the third quadrant and on the line x2 x1 that is y x in more familiar notation The point 1 1 is on the ine x2 x1 but its distance from the origin is J2 So the rotational image of e1 is 12 12 Then this image reflects in the horizontal axis to 1J 1J Similarly e2 rotates into a point on the unit circle that lies in the second quadrant and on the line x2 x1 namely 9 0 H H y n N p n DJ 19 Solutions 63 71x27l2 Then this image re ects in the horizontal axis to AA2M5 When the two calculations described above are written in vertical vector notation the transformation s standard matrix T e1 T e2 is easily seen 1 J5 e gt 1J gt 1J e 1J2 gt 1N A 1J 1 1J 145 2 1J 1A5 1N 15 0 71 elae1 gte2 and eza e2 a el so Ae2 7e11 0 The horizontal shear maps e1 into el and then the re ection in the line 6 7x1 maps e1 into 2 See Table 1 The horizontal shear maps e2 into e2 into e2 7 2e1 To find the image of e2 7 2e1 when it is re ected in the line x2 7x1 use the fact that such a re ection is a linear transformation So the image of e2 7 2e1 is the same linear combination of the images of e2 and e1 namely 1 7 242 7 e1 2e2 To summarize 0 71 e1 gte1 gt e2 and e2 gte2 2e1 gt e12e2 so A 1 2 To nd the image of e 7 2e1 when it is re ected through the vertical axis use the fact that such a re ection is a linear transformation So the image of e2 7 2e1 is the same linear combination of the images of e2 and e1 namely e2 2e1 0 71 0 e1 gt e1 gt e2 and e2 gteZ gt el so A The transformation T described maps e1 gt e1 gt 7e1 and maps e2 gt 7e2 gt 7e2 A rotation through Irradians also maps e1 into 1 and maps e2 into 4 Since a linear transformation is completely determined by what it does to the columns of the identity matrix the rotation transformation has the same effect as T on every vector in R2 The transformation T in Exercise 8 maps e1 gt e1 gt e2 and maps e2 gt 7e2 gt 7e1 Arotation about the origin through 7139 2 radians also maps e1 into e2 and maps e2 into 1 Since a linear transformation is completely determined by what it does to the columns of the identity matrix the rotation transformation has the same effect as T on every vector in R2 Since 2 l 2e1 e2 the image of 2 1 under T is 2Te1 Te2 by linearity of T On the gure in the exercise locate 2T el and use it with T e2 to form the parallelogram shown below 64 14 Equot 9 p A l gt0 N 6 CHAPTER 1 Linear Equations in Linear Algebra Since Tx Ax a1 a2x x1211 Jozaz 7211 3212 when x 71 3 the image ofx is located by forming the parallelogram shown below 3 0 2 x1 3x1 2x3 By inspection 4 0 0 x2 4x1 1 1 1 x3 x1 x2 x3 1 1 x1 x2 By inspection 2 1 x1 2x1 x2 x2 1 0 x1 To express T x as Ax write T x and x as column vectors and then ll in the entries inA by inspection as done in Exercises 15 and 16 Note that since T x and K have four entries A must be a 4gtlt4 matrix 0 x1 0 0 0 0 x1 x x x 1 1 0 0 x 1 2 A 2 0 1 1 0 2 x2 x3 x3 x3 x3 x4 x4 0 0 1 1 x4 As in Exercise 17 write T x and x as column vectors Since x has 2 entriesA has 2 columns Since T x has 4 entriesA has 4 rows 2x2 3x1 3 2 xi4x2 A xi 2 1 4 xi x2 0 0 x2 x2 0 1 Since Tx has 2 entriesA has 2 rows Since x has 3 entries A has 3 columns xl 5x24x3 xl 1 5 4 xl A x2 x2 x2 6 0 1 6 x3 x3 Since Tx has 1 entryA has 1 row Since x has 4 entriesA has 4 columns x1 x1 x2 x2 21513 4x4 A l 2 3 4 21 22 N CA N l The standard matrixA of the transformation T in Exercise 19 is P 19 Solutions 65 11121211 A liHi 1 3 1x1 To solve Tx 8 row reduce the augmented matrix 5 x2 1 l 3 l l 3 0 7 7 l4 s 8H0 1 41 al FM xl sz 1 2 1 Tx x13x2 A x1 1 3 x1To solveTx 4 rowreducethe augmented 399 2x2 x2 3 2 x2 9 matrix 1 2 l l 2 l l 2 l 0 5 1 3 40130130 3xm 3 290412000000 True See Theorem 10 True See Example 3 a b 0 False See the paragraph before Table 1 1 False See the definition of onto Any function from Rquot to R maps each vector onto another vector e False See Example 5 False See the paragraph preceding Example 2 True See Table 1 False See the definition of onetoone Any function from Rquot to R maps a vector onto a single unique vector 21 b True See Theorem 10 c d e True See the solution of Example 5 Three row interchanges on the standard matrixA of the transformation T in Exercise 17 produce This matrix shows thatA has only three pivot positions so the equationAx 0 has a oo 1 0 0 1 0 0 C1 1 0 0 0 0 nontrivial solution By Theorem 11 the transformation T is not onetoone Also sinceA does not have a pivot in each row the columns ofA do not span R4 By Theorem 12 T does not map R4 onto R4 The standard matrixA of the transformation T in Exercise 2 is 2gtlt3 Its columns are linearly dependent becauseA has more columns than rows So T is not onetoone by Theorem 12 Also A is row equivalent to which shows that the rows of A span R2 By Theorem 12 T maps R3 onto R2 5 4 0 D 6 linearly dependent becauseA has more columns than rows So T is not onetoone by Theorem 12 Also A has a pivot in each row so the rows ofA span R2 By Theorem 12 Tmaps R3 onto R2 The columns of A are 66 28 b p A 43 N 43 DJ 43 A CHAPTER 1 Linear Equations in Linear Algebra The standard matrixA of the transformation T in Exercise 14 has linearly independent columns because the gure in that exercise shows that al and 212 are not multiples So T is onetoone by Theorem 12 AlsoA must have a pivot in each column because the equationAx 0 has no free variables Thus the I echelon form ofA is 0 I SinceA has a pivot in each row the columns ofA span R2 So T maps R2 onto R2 An alternate argument for the second part is to observe directly from the gure in Exercise 14 that al and a2 span R2 This is more or less evident based on experience with grids such as those in Figure 8 and Exercise 7 of Section 13 By Theorem 12 the columns of the standard matrixA must be linearly independent and hence the I 0 I equatlon Ax 0 has no free var1ables So each column ofA must be a p1vot column A N 0 0 I 0 0 0 Note that T cannot be onto because of the shape ofA By Theorem 12 the columns of the standard matrixA must span 1R3 By Theorem 4 the matrix must have a pivot in each row There are four possibilities for the echelon form 0 0l0l00l00l 00l 000 000 000 Note that T cannot be onetoone because of the shape ofA T is onetoone if and only ifA has n pivot columns By Theorem 12b T is onetoone if and only if the columns of A are linearly independent And from the statement in Exercise 30 in Section 17 the columns of A are linearly independent if and only if A has n pivot columns The transformation T maps Rquot onto R if and only if the columns of A span R by Theorem 12 This happens if and only if A has a pivot position in each row by Theorem 4 in Section 14 SinceA has m rows this happens if and only ifA has m pivot columns Thus T maps Rquot onto R if and onlyA has m pivot columns De ne T Rquot gt Rm by Tx Bx for some an matrix B and letA be the standard matrix for T By de nitionA T e1 T en where e is the jth column of 11 However by matrixvector multiplication Tej Be b thejth column ofB So A b1 bn B The transformation T maps Rquot onto R if and only if for each y in R there exists an x in Rquot such that y Tx If T Rquot gt Rm maps Rquot onto Rm then its standard matrixA has a pivot in each row by Theorem 12 and by Theorem 4 in Section 14 SoA must have at least as many columns as rows That is m g n When T is onetoone A must have a pivot in each column by Theorem 12 so m 3 n Take u and v in RF and let 0 and dbe scalars Then TScu dv TcASu dSv c39TSu dATSv because T is linear This calculation shows that the mapping x gt T Sx is linear See equation 4 in Section 18 because S is linear M 110 Solutions 67 5 10 5 4 1 0 0 4435 0 0 12571 8 3 4 7 0 1 0 7935 0 0 22571 N111N N There 1s no p1vot 1n the 9 5 3 0 0 1 8635 0 0 D 24571 3 2 5 4 0 0 0 0 0 0 0 0 fourth column of the standard matrixA so the equationAx 0 has a nontrivial solution By Theorem 11 the transformation T is not onetoone For a shorter argument use the result of Exercise 31 Ml 7 5 4 9 0 7 0 10 6 16 4 0 9 0 N 1 11 N No There is no pivot in the third column of the 12 8 12 7 0 0 0 8 6 2 5 0 0 0 0 standard matrixA so the equationAx 0 has a nontrivial solution By Theorem 11 the transformation T is not onetoone For a shorter argument use the result of Exercise 31 4 7 3 7 5 D 0 0 5 0 6 8 5 12 8 0 D 0 1 0 39 M 7 10 8 9 l4 N111N 0 0 D 2 0 There is not a pivot in every row so 3 5 4 2 6 0 0 0 0 D 7 5 6 6 7 37 7 0 0 0 0 07 the columns of the standard matrix do not span R5 By Theorem 12 the transformation T does not map R5 onto R5 9 13 5 6 1 D 0 0 0 5 14 15 7 6 4 0 0 0 4 40 M 8 9 12 5 9 N111N 0 0 0 0 There is not a pivot in every row so 5 6 8 9 8 0 0 0 1 13 14 15 2 1 17 0 0 0 0 0 the coTumns of the standard matrix do not span R5 By Theorem 12 the transformation T does not map R5 onto R5 110 1 a SOLUTIONS If x1 is the number of servings of Cheerios and x2 is the number of servings of 100 Natural Cereal then x1 and x2 should satisfy nutrients nutrients uantities x1 per serving x2 per serving of o nutrients of Cheerios 100 Natural required That is 1 10 13 0 295 4 3 9 x1 x2 20 18 48 68 CHAPTERl Linear Equations in Linear Algebra 110 130 295 4 3 x1 9 b The equ1valent matr1x equatlon 1s To solve thls row reduce the augmented 20 18 x2 48 2 5 8 matrix for this equation 1 10 130 295 2 5 8 1 25 4 4 3 9 4 3 9 4 3 9 20 18 48 N 20 18 48 N 10 9 24 2 5 8 110 130 295 110 130 295 1 25 4 1 25 4 1 0 15 0 7 7 0 1 1 0 1 1 N 16 16 N 0 0 0 N 0 0 0 0 145 145 0 0 0 0 0 0 The desired nutrients are provided by 15 servings of Cheerios together with 1 serving of 100 Natural Cereal 2 Set up nutrient vectors for one serving of Kellogg s Cracklin Oat Bran COB and Kellogg39s Crispix Cm Nutrients COB Crp calories 110 110 protein 3 2 carbohydrate 21 25 fat 3 4 l 10 l 10 3 2 3 aLetBCOB Crp 21 25 u 3 4 Then Bu lists the amounts of calories protein carbohydrate and fat in a mixture of three servings of Cracklin Oat Bran and two servings of Crispix b Let MI and uz be the number of servings of Cracklin Oat Bran and Crispix respectively Can these 110 1 numbers satisfy the equation B 2 225 24 To nd out row reduce the augmented matrix 1 110 110 110 1 1 1 3 2 225 3 2 225 1 0 21 25 24 21 25 24 0 4 3 N 0 0 0 3 4 1 3 4 1 0 26 2 0 0 05 110 Solutions 69 The last row identi es an inconsistent system because 0 705 is impossible So technically there is no mixture of the two cereals that will supply exactly the desired list of nutrients However one could tentatively ignore the nal equation and see what the other equations prescribe They reduce to ul 25 and uz 75 What does the corresponding mixture provide 110 110 110 3 2 225 25 COB 75 C1p 25 75 21 25 24 3 4 105 The error of 5 for fat might be acceptable for practical purposes Actually the data in COB and Crp are certainly not precise and may have some errors even greater than 5 3 Here are the data assembled 39om Table 1 and Exercise 3 Mg of NutrientsUnit Nutrients Nutrient milk flgilr whey 81 fi lgrlg s protein 36 51 13 80 33 carboh 52 34 74 0 45 fat 0 7 11 3 4 3 calcium 126 19 8 18 8 a Let x1 x2 x3 x4 represent the number of units of nonfat milk soy our whey and isolated soy protein respectively These amounts must satisfy the following matrix equation 36 51 13 80 x1 33 52 34 74 0 x2 45 0 7 11 34 3 126 19 8 18 x4 8 36 51 13 80 33 1 0 0 0 64 km 52 34 74 0 45 0 1 0 0 54 0 7 11 34 3 0 0 1 0 09 126 19 8 18 8 0 0 0 1 21 The solution is x1 64 x2 54 x3 709 x4 721 This solution is not feasible because the mixture cannot include negative amounts of whey and isolated soy protein Although the coef cients of these two ingredients are fairly small they cannot be ignored The mixture of 64 units of nonfat milk and 54 units of soy our provide 506 g of protein 516 g of carbohydrate 38 g of fat and 9 g of calcium Some of these nutrients are nowhere close to the desired amounts 4 Let x1 x2 and x3 be the number of units of foods 1 2 and 3 respectively needed for a meal The values of x1 x2 and x3 should satisfy nutrients nvtfients Qtrients milligrams x In mg x 111 mg x 1 mg 2 of nutrients 1 per unlt 2 Per 11m 3 Per mt re uired ofFood l of Food 2 ofFood 3 q 70 CHAPTERI Linear Equations in Linear Algebra From the given data 10 20 20 100 x150 x2 40 963 10 300 30 10 40 200 To solve row reduce the corresponding augmented matrix 10 20 20 100 10 20 20 100 1 2 2 10 50 40 10 300 N 0 60 90 200 N 0 1 32 103 30 10 40 200 0 50 20 100 0 5 2 10 1 2 2 10 1 2 0 25033 1 0 0 5011 N 0 1 32 103 N 0 1 0 5033 N 0 1 0 5033 0 0 1 4033 0 0 1 4033 0 0 1 4033 5011 455 units ofFood1 x 5033 i 152 units ofFood 2 4033 121 units ofFood 3 5 Loop 1 The resistance vector is 5 Total of four RI voltage drops for current 1 r1 2 Voltage drop for 12 is negative 12 ows in opposite direction 0 Current 13 does not ow in loop 1 0 Current 14 does not ow in loop 1 Loop 2 The resistance vector is 2 Voltage drop for 11 is negative I1 ows in opposite direction r 11 Total of four RI voltage drops for current 12 2 3 Voltage drop for 13 is negative I 3 ows in opposite direction 0 Current 14 does not ow in loop 2 0 0 5 2 0 3 0 Also r3 r4 andR r1 r2 r3 r4 2 11 3 0 17 4 0 3 17 4 4 25 0 0 4 25 Notice that each offdiagonal entry of R is negative or zero This happens because the loop current directions are all chosen in the same direction on the gure For each loop j this choice forces the currents in other loops adjacent to loop j to ow in the direction opposite to currenth 40 Next set v 30 The voltages in loops 2 and 4 are negative because the battery orientation in each 20 10 loop is opposite to the direction chosen for positive current ow Thus the equation Ri v becomes 5 2 2 11 0 3 0 0 110 Solutions 71 0 11 40 11 756 1 1 3 2 30 MThe solution isi 2 13910 17 4 13 20 13 93 4 25 14 10 14 25 6 Loop 1 The resistance vector is l Total of four RI voltage drops for current 11 Voltage drop for 12 is negative 12 ows in opposite direction Current 13 does not ow in loop 1 Current 14 does not ow in loop 1 Loop 2 The resistance vector is 1 r2 6 2 0 Also r3 4 1 1 6 0 2 0 0 Voltage drop for 11 is negative 11 ows in opposite direction Total of four RI voltage drops for current 12 Voltage drop for I 3 is negative 13 ows in opposite direction Current 14 does not ow in loop 2 0 0 40 2 dR St 30 Th Rquot b 10 r 3 an r1 r2 r3 r e v 0 en 1 v ecomes 3 12 10 0 0 11 40 11 1211 2 0 I 30 I 844 2 M The solution is i 2 10 3 I3 20 13 426 3 12 I4 10 14 190 Loop 1 The resistance vector is Total of three RI voltage drops for current 1 Voltage drop for 12 is negative 12 ows in opposite direction Current 13 does not ow in loop 1 Voltage drop for 14 is negative 14 ows in opposite direction Loop 2 The resistance vector is Voltage drop for 11 is negative 11 ows in opposite direction Total of three RI voltage drops for current 12 Voltage drop for 13 is negative 13 ows in opposite direction Current 14 does not ow in loop 2 72 CHAPTERI Linear Equations in Linear Algebra 0 4 12 7 0 4 6 0 7 15 6 0 Alsor r andR r r r r 3 14 5 1 2 3 4 0 6 14 5 5 13 4 0 5 13 Notice that each offdiagonal entry of R is negative or zero This happens because the loop current directions are all chosen in the same direction on the gure For each loop j this choice forces the currents in other loops adjacent to loop j to ow in the direction opposite to currenth 40 30 Next set v 20 Note the negatlve voltage 111 loop 4 The current d1rectlon chosen in loop 4 1s 10 opposed by the orientation of the voltage source in that loop Thus Ri v becomes 12 7 0 4 11 40 11 1143 7 15 6 0 2 30 12 1055 M The solutlon 1s 1 0 6 14 5 I3 20 13 804 4 0 5 13 I4 10 14 584 8 Loop 1 The resistance vector is 15 Total of four RI voltage drops for current 11 5 Voltage drop for 12 is negative 12 ows in opposite direction r1 0 Current 3 does not ow in loop 1 5 Voltage drop for 14 is negative 14 ows in opposite direction 1 Voltage drop for 15 is negative 15 ows in opposite direction Loop 2 The resistance vector is 5 Voltage drop for 11 is negative 11 ows in opposite direction 15 Total of four RI voltage drops for current 12 r2 5 Voltage drop for 13 is negative 13 ows in opposite direction 0 Current 14 does not ow in loop 2 2 Voltage drop for 15 is negative 15 ows in opposite direction 0 5 1 15 5 0 5 1 40 5 0 2 5 15 5 0 2 30 Also r3 15 r4 5 r5 3 andR 0 5 15 5 3 Setv 20 Note the 5 15 4 5 0 5 15 4 10 3 4 10 1 2 3 4 10 0 negative voltages for loops where the chosen current direction is opposed by the orientation of the voltage source in that loop Thus Ri v becomes H H 110 Solutions 73 15 5 0 5 1 11 40 11 337 5 15 5 0 2 12 30 12 11 0 5 15 5 3 I3 20 M The solution is 13 227 5 0 5 15 4 I4 10 14 167 1 2 3 4 10 5 0 15 170 9 The population movement problems in this section assume that the total population is constant with no migratlon or 1mm1grat10n The statement that about 5 of the city s population moves to the suburbs means also that the rest of the city s population 95 remain in the city This determines the entries in the rst column of the migration matrix which concerns movement from the city From City Suburbs To 95 City 05 Suburbs Likewise if 4 of the suburban population moves to the city then the other 96 remain in the suburbs 95 04 This determines the second column of the migration matrix M 05 96 J The difference equation is xk Mxk for k 0 1 2 Also x0 400 000 95 04 600000 586000 The populatlon m 2001 when k 1 1s x1 Mx0 05 600 000 96 400 000 414 000 95 04 586000 573260 The populatlon m 2002 when k 2 1s X2 Mx1 05 96 414000 426740 The data in the rst sentence implies that the migration matrix has the form From City Suburbs To 03 City 07 Suburbs The remaining entries are determined by the fact that the numbers in each column must sum to 1 For instance if 7 of the city people move to the suburbs then the rest or 93 remain in the city So the 93 03 800 000 m1grat10n matrlx 1sM The m1t1al populatlon 1s x0 07 97 500 000 93 03 800000 759000 The populatlon m 2001 when k 1 1s X1 Mxo 07 97 500000 541000 93 03 759 000 722100 The populatlon m 2002 when k 2 1s x2 Mxl 07 97 541 000 577900 The problem concerns two groups of peopleithose living in California and those living outside California and in the United States It is reasonable but not essential to consider the people living inside 74 CHAPTERI Linear Equations in Linear Algebra California rst That is the first entry in a column or row of a vector will concern the people living in California With this choice the migration matrix has the form From Calif Outside To Calif Outside a For the rst column of the migration matrix M compute Calif persons who moved 509 500 Total Calif pop 29726000 The other entry in the rst column is 1 7 017146 982854 The exercise requests that 5 decimal places he used So this number should be rounded to 98285 Whatever number of decimal places is used it is important that the two entries sum to 1 So for the rst fraction use 01715 017146 outside personsl who moved J 564100 Total outside pop 2 218994000 is 1 7 00258 99742 Thus the migration matrix is From Calif Outside To 98285 00258 Calif For the second column of M compute 00258 The other entry 01715 99742 Outside b M The initial vector is x0 29716 218994 with data in millions of persons Since x0 describes the population in 1990 and x1 describes the population in 1991 the vector x10 describes the projected population for the year 2000 assuming that the migration rates remain constant and there are no deaths births or migration Here are some of the vectors in the calculation with only the rst 4 or 5 gures displayed Numbers are in millions of persons 297 298 298 301 3018 30223 7 2190 2189 2189 2186 21853 218487 X10 97 05 10 305 97 05 10 305 308 12 SetM 00 90 05 and x0 48 Then x1 00 90 05 48 z 48 and 03 05 85 98 03 05 85 98 95 97 05 10 308 311 X2 00 90 05 48 2 48 The entries in X2 give the approximate distribution of cars on 03 05 85 95 92 Wednesday two days after Monday S43 M The order of entries in a column of a migration matrix must match the order of the columns For instance if the rst column concerns the population in the city then the rst entry in each column must be the fraction of the population that moves to or remains in the city In this case the data in the 95 03 600000 exerclse leads to M and x0 05 97 400000 110 Solutions 75 21 Some of the population vectors are 523293 472737 439417 417456 X5 X10 X15 X20 2 476 707 527 263 560 583 582 544 The data here shows that the city population is declining and the suburban population is increasing but the changes in population each year seem to grow smaller 3 50 000 b When x0 the situation is different Now 650000 358 523 364140 367 843 370 283 X5 X10 X15 X20 2 641477 635860 632157 629717 The city population is increasing slowly and the suburban population is decreasing No other conclusions are expected This example will be analyzed in greater detail later in the text 14 Here are Figs a and b for Exercise 13 followed by the gure for Exercise 34 in Section 11 20 2 20 2 0 1 2 0 10 1 2 4A 10 1 2 4A 0 4 3 10 4 3 40 10 4 3 4 20 2 1 1 30 3 3 b Section 11 For Fig a the equations are 4T1 020T2 T4 4T2 T1200T3 4T3 T4T2020 4T4 0T1T3 20 To solve the system rearrange the equations and row reduce the augmented matrix Interchanging rows 1 and 4 speeds up the calculations The rst ve steps are shown in detail 4 1 0 1 20 1 0 1 4 20 1 0 1 4 20 1 0 1 4 20 1 4 1 0 20 1 4 1 0 20 0 4 0 4 0 0 1 0 1 0 0 1 4 1 20 N 0 1 4 1 20 N 0 1 4 1 20 N 0 1 4 1 20 1 0 1 4 20 4 1 0 76 CHAPTER 1 Linear Equations in Linear Algebra For Fig b the equations are 4T1 100T2T4 4T2 T1040T3 4T3 T4T24010 4T4 10T1T310 Rearrange the equations and row reduce the augmented matrix 4 71 0 71 10 1 0 0 0 10 71 4 71 0 40 0 1 0 0 175 0 71 4 71 50 Ni N 0 0 1 0 20 71 0 71 4 20 0 0 0 1 125 a P O Here are the solution temperatures for the three problems studied Fig a in Exercise 14 of Section 110 10 10 10 10 Fig b in Exercise 14 of Section 110 10 175 20 125 Figure for Exercises 34 in Section 11 20 275 30 225 When the solutions are arranged this way it is evident that the third solution is the sum of the rst two solutions What might not be so evident is that list of boundary temperatures of the third problem is the sum of the lists of boundary temperatures of the rst two problems The temperatures are listed clockwise starting at the left of T1 Fig a 0 20 20 0 0 20 20 0 Fig b 10 0 0 40 40 10 10 10 Fig from Section 11 10 20 20 40 40 30 30 10 When the boundary temperatures in Fig a are multiplied by 3 the new interior temperatures are also multiplied by 3 The correspondence from the list of eight boundary temperatures to the list of four interior temper atures is a linear transformation A veri cation of this statement is not expected However it can be shown that the solutions of the steadystate temperature problem here satisfy a superposition principle The system of equations that approximate the interior temperatures can be written in the formAx b whereA is determined by the arrangement of the four interior points on the plate and b is a vector in R4 determined by the boundary temperatures Note The MATLAB box in the Study Guide for Section 110 discusses scienti c notation and shows how to generate a matrix whose columns list the vectors x0 x1 x2 determined by an equation xk Mxk for k 0 1 Chapter 1 SUPPLEMENTARY EXERCISES 1 21 False The word reduced is missing Counterexample Al 3 31643 ll The matrixA is row equivalent to matrices B and C both in echelon form 0 Q 5quot 1quoti IO 5quot r 5 O quot5 quot3D Chapterl Supplementary Exercises 77 False Counterexample LetA be any an matrix with fewer than n pivot columns Then the equation Ax 0 has in nitely many solutions Theorem 2 in Section 12 says that a system has either zero one or infinitely many solutions but it does not say that a system with in nitely many solutions exists Some counterexample is needed True If a linear system has more than one solution it is a consistent system and has a free variable By the Existence and Uniqueness Theorem in Section 12 the system has in nitely many solutions False Counterexample The following system has no free variables and no solution x1 x2 1 x2 5 x1 x2 2 True See the box after the de nition of elementary row operations in Section 11 If A b is transformed into C d by elementary row operations then the two augmented matrices are row equivalent True Theorem 6 in Section 15 essentially says that whenAx b is consistent the solution sets of the nonhomogeneous equation and the homogeneous equation are translates of each other In this case the two equations have the same number of solutions False For the columns of A to span R39quot the equationAx b must be consistent for all h in R not for just one vector b in Rm False Any matrix can be transformed by elementary row operations into reduced echelon form but not every matrix equation Ax b is consistent True IfA is row equivalent to B thenA can be transformed by elementary row operations rst into B and then further transformed into the reduced echelon form U of B Since the reduced echelon form of A is unique it must be U False Every equationAx 0 has the trivial solution whether or not some variables are free True by Theorem 4 in Section 14 If the equationAx b is consistent for every b in Rm thenA must have a position in every one of its m rows If A has m pivot positions thenA has m pivot columns each containing one pivot position False The word unique should be deleted LetA be any matrix with m pivot columns but more than m columns altogether Then the equationAx b is consistent and has m basic variables and at least one free variable Thus the equation does not does not have a unique solution True If A has n pivot positions it has a pivot in each of its n columns and in each of its n rows The reduced echelon form has a 1 in each pivot position so the reduced echelon form is the an identity matrix True Both matrices A and B can be row reduced to the 3X3 identity matrix as discussed in the previous question Since the row operations that transform B into 13 are reversible A can be transformed rst into 13 and then into B True The reason is essentially the same as that given for question f True If the columns of A span R then the reduced echelon form of A is a matrix U with a pivot in each row by Theorem 4 in Section 14 Since B is row equivalent to A B can be transformed by row operations rst into A and then further transformed into U Since U has a pivot in each row so does B By Theorem 4 the columns of B span Rm False See Example 5 in Section 16 True Any set of three vectors in R2 would have to be linearly dependent by Theorem 8 in Section 16 78 CHAPTERl Linear Equations in Linear Algebra 5 False Ifa set v1 v2 v3 v4 were to span R5 then the matrixA v1 v2 v3 v4 would have a pivot position in each of its five rows which is impossible sinceA has only four columns t True The vector 7u is a linear combination of u and v namely iu 71u 0v u False If u and v are multiples then Spanu v is a line and w need not be on that line v False Let u and v be any linearly independent pair of vectors and let w 2v Then w 0H 2v so w is a linear combination of u and v However u cannot be a linear combination of v and w because if it were u would be a multiple of v That is not possible since u v is linearly independent w False The statement would be true if the condition v1 is not zero were present See Theorem 7 in Section 17 However if v1 0 then v1 v2 v3 is linearly dependent no matter what else might be true about v2 and v3 x True Function is another word used for 1ransformation as mentioned in the de nition of 1ransformation in Section 18 and a linear transformation is a special type of transformation y True For the transformation x gtAx to map R5 onto R6 the matrixA would have to have a pivot in every row and hence have six pivot columns This is impossible becauseA has only five columns z False For the transformation x gtAx to be onetoone A must have a pivot in each column Since A has n columns and m pivots m might be less than n 2 Ifa 0 then x ba the solution is unique Ifa 0 and b 0 the solution set is empty because 0x 0 b If a 0 and b 0 the equation 0x 0 has in nitely many solutions 3 a Any consistent linear system whose echelon form is 0lor00lor00l 0 0 0 0 0 0 0 0 0 0 0 0 b Any consistent linear system whose coefficient matrix has reduced echelon form 13 c Any inconsistent linear system of three equations in three variables 4 Since there are three pivots one in each row the augmented matrix must reduce to the form I 0 l A solution ofo b exists for all h because there is a pivot in each row ofA Each 0 0 l solution is unique because there are no free variables 1 3 k 3 k 5 a N Ifh 12 and k i 2 the second row of the augmented matrix 4 h 8 0 h 12 8 4k indicates an inconsistent system of the form 0 b with b nonzero If h 12 and k 2 there is only one nonzero equation and the system has infmitely many solutions Finally if h i 12 the coefficient matrix has two pivots and the system has a unique solution 9 2 h 1 h 1 N If k 311 0 the system 1s 1ncons1stent Otherwrse the 6 k 2 0 k 311 1 coef cient matrix has two pivots and the system has a unique solution Chapterl Supplementary Exercises 79 4 2 7 5 6 a Set v1 8 v2 3 v3 10 and b 3 Determine ifb is a linear combination ofvl v2 v3 Or Determine if b is in Spanv1 v2 v3 To do this compute 4 27 5 27 5Th t tt bs N es semlscons1sen so zsm anvvv 8 310 30 47 y P123 4 2 7 5 b SetA 8 3 10 b 3 Determlne 1fb 1s a linear combinatlon ofthe columns ofA c De ne Tx Ax Determine if b is in the range of T 2 4 2 b1 7 a Set v1 5 v2 l v3 l and b b2 Determine ifvl v2 v3 spanR3 To do this row 7 5 3 b3 reduce v1 v2 v3 2 4 2 2 4 2 4 2 5 l l N 0 9 4 N 0 4 The matrix does not have a pivot in each row so 7 5 3 0 9 4 0 0 0 its columns do not span R3 by Theorem 4 in Section 14 2 4 2 b SetA 5 l l Determine ifthe columns ofA span R3 7 5 3 c De ne T x Ax Determine if T maps R3 onto R3 I 8 a b 0 l 0 l 0 0 l 0 0 l 0 0 l 1 2 9 The rst line 1s the hue spanned by 2 The second 11ne 1s spanned by I So the problem 1s to wr1te 5 1 2 6 as the sum of a multlple of 2 and a multlple of 1 That 1s nd x1 and x2 such that 2 1 5 x1 I x2 2 6 Reduce the augmented matrlx for thls equatlon 2 1 5 1 2 6 1 2 6 1 2 6 1 0 4 3 1 2 6 2 1 5 0 3 7 0 1 73 0 1 73 5 4 2 7 l 5 8 3 7 3 Thus or 6 3 1 3 2 6 4 3 14 3 10 The line through 211 and the origin and the line through 212 and the origin determine a grid on the xlxzplane as shown below Every point in R2 can be described uniquely in terms of this grid Thus b can 80 CHAPTERl Linear Equations in Linear Algebra be reached from the origin by traveling a certain number of units in the aldirection and a certain number of units in the azdirection H H A solution set is a line when the system has one free variable If the coefficient matrix is 2gtlt3 then two of 2 0 3 4 Put anything in column 3 The resulting matrix will be in echelon form Make one row replacement operation on the second row to l the columns should be pivot columns For instance take l 2 l l 2 1 create a matr1x not 1n echelon form such as N 0 3 l l 5 2 N A solution set is a plane where there are two free variables If the coefficient matrix is 2gtlt3 then only one column can be a pivot column The echelon form will have all zeros in the second row Use a row 1 2 3 replacement to create a matrix not in echelon form For instance let A 1 2 3 gtllt p A DJ The reduced echelon form ofA looks like E 0 Since E is row equivalent toA the equation 0 l 0 l 0 l 0 Ex 0 has the same solutions as Ax 0 Thus 0 l 0 0 l 0 3 By inspection E 0 l 0 0 0 p A A l 0 Row reduce the augmented matrix for x1 x2 a a a 2 0 l a 0 1 a 0 l a 0 a 12 0 0 a2 a2 0 0 Z ala 0 The equation has a nontrivial solution only when 2 7 al a 0 So the vectors are linearly independent for all 1 except a 2 and a 71 15 a If the three vectors are linearly independent then a c and f must all be nonzero The converse is true too LetA be the matrix whose columns are the three linearly independent vectors Then p A Ch p A l 9 p A D 22 Denote the columns from right to left by v1 Chapter 1 Supplementary Exercises 81 A must have three pivot columns See Exercise 30 in Section 17 or realize that the equationAx 0 has only the trivial solution and so there can be no free variables in the system of equations Since A is 3 X3 the pivot positions are exactly where a c and f are located b The numbers a f can have any values Here s why Denote the colunms by v1 v2 and v3 Observe that v1 is not the zero vector Next v2 is not a multiple of v1 because the third entry of v2 is nonzero Finally v3 is not a linear combination of v1 and v2 because the fourth entry of v3 is nonzero By Theorem 7 in Section 17 v1 v2 v3 is linearly independent v4 The first vector v1 is nonzero v2 is not a multiple of VI because the third entry of v2 is nonzero and v3 is not a linear combination of VI and v2 because the second entry of v3 is nonzero Finally by looking at rst entries in the vectors v4 cannot be a linear combination of v1 v2 and v3 By Theorem 7 in Section 17 the columns are linearly independent Here are two arguments The rst is a direct proof The second is called a proof by contradiction H Since v1 v2 v3 is a linearly independent set v1 0 Also Theorem 7 shows that v2 cannot be a multiple of VI and v3 cannot be a linear combination of VI and v2 By hypothesis v4 is not a linear combination of v1 v2 and v3 Thus by Theorem 7 v1 v2 v3 v4 cannot be a linearly dependent set and so must be linearly independent Suppose that v1 v2 v3 v4 is linearly dependent Then by Theorem 7 one of the vectors in the set is a linear combination of the preceding vectors This vector cannot be v4 because v4 is not in Spanv1 v2 v3 Also none of the vectors in v1 v2 v3 is a linear combinations of the preceding vectors by Theorem 7 So the linear dependence of v1 v2 v3 v4 is impossible Thus v1 v2 v3 v4 is linearly independent Suppose that cl and C are constants such that Then 01 czv1 czvz 0 Since v1 and v2 are linearly independent both cl C 0 and C 0 It follows that both cl and C in must be zero which shows that v1 v1 v2 is linearly independent cm 02V1 V2 0 LetM be the line through the origin that is parallel to the line through v1 v2 and v3 Then v2 7 v1 and v3 7 v1 are both onM So one of these two vectors is a multiple ofthe other say v2 7 v1 kv3 7 v1 This equation produces a linear dependence relation k 7 lv1 v2 7 kv3 0 A second solution A parametric equation of the line is x v1 tv2 7 v1 Since v3 is on the line there is some to such that v3 v1 t0v2 7v1 l 7 t0v1 tovz So v3 is a linear combination of VI and v2 and v1 v2 v3 is linearly dependent IfTu v then since T is linear TH T1u 1Tu V Either compute T e1 T e2 and T e3 to make the colunms of A or write the vectors vertically in the de nition of T and ll in the entries of A by inspection x1 x1 1 0 0 Ax A x2 A 0 71 0 0 0 1 x2 x3 x3 By Theorem 12 in Section 19 the colunms of A span R3 By Theorem 4 in Section 14A has a pivot in each of its three rows SinceA has three colunms each colunm must be a pivot colunm So the equation 82 CHAPTERl Linear Equations in Linear Algebra Ax 0 has no free variables and the columns of A are linearly independent By Theorem 12 in Section 19 the transformation x gtAx is onetoone a b 4 5 4a 3b 5 23 1mphes that Solve b a 3 0 311 4b 0 4 3 5 4 3 5 4 3 5 4 0 165 1 0 45 3 4 0 0 254 154 0 1 35 0 1 35 0 1 35 Thus a 45 and b 735 24 The matrix equation displayed gives the information 211 4b 2J5 and 411 2b 0 Solve for a and b 242J N2 4 2J N1 2 J5N101x5 4 2 0 010 4J 1 2J3 1 2J3 So a1J b 2J 25 a The vector lists the number of three two and onebedroom apartments provided when x1 oors of planA are constructed 3 4 5 bx17x24g3 8 8 9 3 4 5 66 c M Solvex1 7 x2 4 x3 3 74 8 8 9 136 3 4 5 66 1 0 12 2 x1 12x3 2 7 4 3 74 NM 0 1 138 15 x2 138x3 15 8 8 9 136 0 0 0 0 0 0 The general solution is x1 212g 2 12 x x2 15 138g 15 Jg 138 x3 x3 0 1 However the only feasible solutions must have whole numbers of oors for each plan Thus 03 must be a multiple of 8 to avoid fractions One solution for x3 0 is to use 2 oors of plan A and 15 oors of plan B Another solution for x3 8 is to use 6 oors of plan A 2 oors of plan B and 8 oors of plan C These are the only feasible solutions A larger positive multiple of 8 for x3 makes x2 negative A negative value for x3 of course is not feasible either Determinants 31 SOLUTIONS Notes Some exercises in this section provide practice in computing determinants while others allow the student to discover the properties of determinants which will be studied in the next section Determinants are developed through the cofactor expansion which is given in Theorem 1 Exercises 33 36 in this section provide the first step in the inductive proof of Theorem 3 in the next section 1 Expanding along the first row 3 O 4 3 2 2 2 2 3 2 3 2 3 4 3 134101 5 1 O 1 O 5 O 5 1 Expanding along the second column 3 O 4 2 2 3 4 3 4 2 3 2 1120 1223 1325 3 3 5 2 1 ltgt01ltgt01ltgt22ltgtltgt O 5 1 2 Expanding along the first row 0 5 1 3 O 4 O 4 3 4 3 0 0 5 1 541222 4 1 2 1 2 4 2 4 1 Expanding along the second column 0 5 1 4 O O 1 O 1 4 3 0 1125 122 3 1324 2 54 3 2 4 4 2 2412121 40 3 Expanding along the first row 2 4 3 3 1 2 21 2 4 3 2 3 3 1 2 94 5311 5 1 4 1 4 1 1 1 1 4 159 160 4 UI ex 1 8 9 CHAPTER3 Determinants Expanding along the second column 2 4 3 3 2 2 3 2 3 3 1 2 1 2 4 12 1 132 4 4 5 1 5 4 5 5 141 111132 Expanding along the first row l 3 5 2 1 11 1 1 3 2 152 1 1 2 315520 3 4 2 2 3 2 3 4 Expanding along the second column 1 3 5 12 2 1 22 1 5 32 1 5 i l 3 3 2 1 1 3 2 1 4 2 1 311 13 4 920 Expanding along the first row 2 3 4 4 0 5 20 5 34 544 0 25 31 44 23 51 16 56 51 Expanding along the first row 5 2 4 0 3 5 5 3 5 20 5 40 3 5121046 1 4 7 2 7 2 4 2 4 7 Expanding along the first row 4 3 0 5 2 6 2 6 5 6 5 24 3 0 41 304 7 3 9 3 9 7 9 7 3 Expanding along the first row 816 4 0 3 80 314 3 64 0 86 111 68 11 325 25 35 3 2 First expand along the third row then expand along the rst row of the remaining matrix 6 0 0 5 172 5 00 5 72 13 27 2 52 1 35 10110 2 0 0 0 31 318 8 318 31 Solutions 161 10 First expand along the second row then expand along either the third row or the second column of the 1 1 1 1 1 remaining matr1x 1 2 5 2 1 2 2 g 5 123A32 6 5 5 0 4 4 5 0 4 2 2 1 2 3 13 55396 5 1334 2 6U 3524 2 6 01 1 2 5 2 1 2 2 g 5 123A32 6 5 5 0 4 4 5 0 4 12 2 5 3 1 3 5 4 1 2 01 4 6 5 4 32 17 6 6 6 There are many ways to do this determinant efficiently One strategy is to always expand along the rst column of each matrix 3 5 8 4 0 2 7 2 3 7 15 1 13 0 1 53 1 1 2 3722712 0 0 5 0 2 X 0 0 2 0 0 0 2 There are many ways to do this determinant efficiently One strategy is to always expand along the rst row of each matrix 4 0 0 0 7 10 0 1 0 0 3 0 1 1 46 3 04 1 1 1 4717936 2630 43 8 4 3 5 8 4 3 First expand along either the second row or the second column Using the second row 40 73 5 40 3 5 00200 734 8 73 6 4 8 1 2 50 2 3 50 52 3 00 12 00 9 12 Now expand along the second column to nd 4 0 3 5 7 3 4 8 4 3 5 1232 2 12235 2 3 5 0 2 3 0 0 1 2 0 1 2 162 CHAPTER3 Determinants Now expand along either the first column or third row The rst column is used below 4 3 5 2 3 3 5 2 l2213 5 2 3 6 l1114 1 2 12115 1 2U 641 5l6 0 1 2 14 First expand along either the fourth row or the h column Using the fifth column 6 3 2 4 0 6 3 2 4 9 0 4 1 0 35 9 0 4 1 8 5 6 7 l 1 Al 3 0 0 0 3 0 0 0 0 4 2 3 2 4 2 3 2 0 Now expand along the third row to find 6 3 2 4 9 0 4 l 3 2 4 13s1 1 131130 4 1 3 0 0 0 2 3 2 4 2 3 2 Now expand along either the first column or second row The rst column is used below 3 2 4 31 11 4 1 31 2 4 l 1 A3 0 4 1 3 1 A3 3 2 l A2 4 1 33 ll2189 2 3 2 3 0 4 15 2 3 2 331 t 020 t 425i034523120 0 5 1 790407073041 0 5 l 16 4 3 0 031502t144231400145 2 4 l 001674707202 2 4 3 17 3 1 2 211421t334113422134 4 1 7278367371671275 l 3 5 18 2 1 1 112313t524315411223 3 4 2 29407157471220 N p A 23 24 3 31 Solutions 163 a b d The row operation swaps rows 1 and 2 of the matrix and the sign of the determinant is reversed C d ad bc b cb da ad bc a bzadbc a b d kc kd The row operation scales row 2 by k and the determinant is multiplied by k akd kcb kad kbc kad be 36 4k 5 3k4 2 4 18 20 2 5 6 3 53k 64ki The row operation replaces row 2 with k times row 1 plus row 2 and the determinant is unchanged at b ad bc d akc bkd akcd cbkdadkcd bc kcdad bc The row operation replaces row 1 with k times row 2 plus row 1 and the determinant is unchanged 1 1 1 k k k 3 8 4 14 12 1 7 5 3 8 4 k4 k2 k 7 5k 2 3 2 2 3 2 The row operation scales row 1 by k and the determinant is multiplied by k a b c 3 2 2 a2 b6c32a 6b3c 6 5 6 3 2 2 a b c 36b 50 26a 6c25a 6b 2a6b 3c 6 5 6 The row operation swaps rows 1 and 2 of the matrix and the sign of the determinant is reversed Since the matrix is triangular by Theorem 2 the determinant is the product of the diagonal entries 1 0 0 0 1 0 111 1 0 k 1 Since the matrix is triangular by Theorem 2 the determinant is the product of the diagonal entries 1 0 0 0 1 0 111 1 k 0 1 Since the matrix is triangular by Theorem 2 the determinant is the product of the diagonal entries k 0 0 0 1 0 0 0 1 k11 k DJ 03 P Equot CHAPTER3 Determinants Since the matrix is triangular by Theorem 2 the determinant is the product of the diagonal entries 1 0 0 0 k 0 1k1k 0 0 1 A cofactor expansion along row 1 gives 0 l 0 1 0 1 0 0 71 71 0 1 0 0 1 A cofactor expansion along row 1 gives 0 0 1 0 1 0 1 0 1 71 1 0 1 0 0 A 3 X 3 elementary row replacement matrix looks like one of the six matrices 710010010010010k1k0 k1001001001k010010 7001k010k1001001001 In each of these cases the matrix is triangular and its determinant is the product of its diagonal entries which is 1 Thus the determinant of a 3 X 3 elementary row replacement matrix is l A 3 X 3 elementary scaling matrix with k on the diagonal looks like one of the three matrices k 0 0 1 0 0 1 0 0 0 l 0 0 k 0 0 1 0 0 0 1 0 0 l 0 0 k In each of these cases the matrix is triangular and its determinant is the product of its diagonal entries which is k Thus the determinant of a 3 X 3 elementary scaling matrix with k on the diagonal is k 0 b 1 detE 7l detA ad7 be detEA eb 7 da 7lad7 be det EdetA l 0 b a Elo W 3 allele l detE k detA ad7 be det EA akd 7 keb kad 7 be det EdetA bl l l EA d l k a E A 0 l e detE l detA ad7 be detEA a ked7 eb kd ad ked7 be 7ked 1ad7 be det EdetA b kd bkd d ake C DJ 5 DJ 1 DJ on J p A A N a b ka A kA e d Ice 3 The area of the parallelogram determined by u 31 Solutions 165 l 0 a b a b E A EA k l e d kae kbd detE l detA adi be detEA akb d 7 ka eb kab adi kab 7 be 1ad7 be det EdetA 3 l 15 5 A 5A 10 detA2det5A50 5detA 42 20 kb detA adi be kd detkA kakd kbke k2ad be kzdetA a True See the paragraph preceding the de nition of the determinant b False See the definition of cofactor which precedes Theorem 1 21 False See Theorem 1 b False See Theorem 2 v u v and 0 is 6 since the base ofthe l 0 2 parallelogram has length 3 and the height of the parallelogram is 2 By the same reasoning the area of 3 the parallelogram determined by u 0 x 926 u x and 0 is also 6 X1 I 1 2 U 4 3 l 3 x Also note that detu vdet 0 2 6 and detu x det 0 2 6 The determmant of the matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0 is equal to the area of the parallelogram The area of the parallelogram determined by u v u v and 0 is eb since the base of the parallelogram has length c and the height of the parallelogram is b X2 166 32 CHAPTER3 Determinants Also note that detu v detE cb and detv u deth Z Cb The determinant ofthe matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0 either is equal to the area of the parallelogram or is equal to the negative of the area of the parallelogram M Answers will vary The conclusion should be that det A B detA det B M Answers will vary The conclusion should be that det AB detAdet B M Answers will vary For 4 X 4 matrices the conclusions should be that detAT detA detiA detA detZA l6det A and detlOA 104 detA For 5 x 5 matrices the conclusions should be that detAT detA detiA AietA det2A 32detA and detlOA105detA For 6 x 6 matrices the conclusions should be that detAT detA detiA detA detZA 64detA and det 10A 106 detA M Answers will vary The conclusion should be that detA 1 l detA SOLUTIONS Notes This section presents the main properties of the determinant including the effects of row operations on the determinant of a matrix These properties are first studied by examples in Exercises 1720 The properties are treated in a more theoretical manner in later exercises An efficient method for computing the determinant using row reduction and selective cofactor expansion is presented in this section and used in Exercises 11714 Theorems 4 and 6 are used extensively in Chapter 5 The linearity property of the determinant studied in the text is optional but is used in more advanced courses 1 2 3 4 Rows l and 2 are interchanged so the determinant changes sign Theorem 3b The constant 2 may be factored out of the Row 1 Theorem 3c The row replacement operation does not change the determinant Theorem 3a The row replacement operation does not change the determinant Theorem 3a 15 615 615 6 1 4 401 201 23 2 7903 3003 15 315 315 315 3 3 3 30 181260 3 260 3 26 3 18 213 703 103 1001 1302130213021302 2 5740178017801780 35210 42 5003027003027 1 12 30 42 50030270000 1 N 32 Solutions 167 1 3 3 4 13 3 4 13 3 4 0 12 50 12 5012 50 2 5 4 3 0 1 2 5 00 0 0 3 7 5 2 0 2 4 10 0 0 0 0 1 1 301 1 301 1 30 1 1 30 0 1 5 40 1 5 40 1 5 40 1 5 33 1 2 8 5 0 15 5 0 0 0 1 0 0 3 5 3 1 23027300 3 5 0001 1 3 1 0 2 1 3 1 0 2 1 3 1 0 2 02 4 1 602 4 1 602 4 1 6 2 6 2 3 90 0 0 3 50 0 0 3 5 3 7 3 8 7 0 2 0 8 1 0 0 4 7 7 3 5 5 2 7 0 4 8 213 0 0 0 0 1 First use a row replacement to create zeros in the second column and then expand down the second column 2 5 3 l 2 5 3 l 3 l 3 3 0 1 3 3 0 1 3 5 6 4 9 6 0 4 9 6 0 4 9 0 2 1 4 10 4 1 0 0 2 1 Now use a row replacement to create zeros in the rst column and then expand down the rst column 3 1 3 3 1 3 5 6 4 9 5 0 2 3 53 31 53 8 120 0 2 1 0 2 First use a row replacement to create zeros in the fourth column and then expand down the fourth column 1 2 3 0 1 2 3 0 1 2 3 3 4 3 0 3 4 3 0 3 3 4 3 5 4 6 6 3 0 2 0 3 0 2 4 2 4 3 4 2 4 3 Now use a row replacement to create zeros in the rst column and then expand down the rst column 1 2 3 l 10 12 3 3 4 3 3 0 10 12 3 l 6 11 3 0 2 0 6 11 3 1 38 114 168 CHAPTER3 Determinants 13 First use a row replacement to create zeros in the fourth column and then expand down the fourth p A A column 2 5 4 1 2 5 4 1 0 3 2 4 7 6 2 0 3 2 0 1 6 2 4 6 2 4 0 6 2 4 0 6 7 7 6 7 7 0 6 7 7 0 Now use a row replacement to create zeros in the rst column and then expand down the rst column 0 3 2 0 3 2 3 2 1 6 2 4 l 6 2 4 l 6 5 3 1 61 6 6 7 7 0 5 3 First use a row replacement to create zeros in the third column and then expand down the third column 3 2 l 4 3 2 1 4 l 3 3 1 3 0 3 1 3 0 3 1 9 0 0 3 4 2 8 9 0 0 0 3 4 4 3 4 0 4 3 4 0 4 Now expand along the second row 1 3 3 3 3 19 0 0 19 4 4 1900 3 4 4 a b c a b c d e f 5 d e f 5735 5g 5h 5139 g h i a b c a b 0 3d 38 3f 3 d e f 3721 g h i g h i b c a b c g h z d e f 7 d e f g h i g h i b c a c a b c g h i d f 77 d e f d e f g i a b c a b c a b c 2da 2eb 2fc 2d 28 2f 2d e f 27l4 g h i g h i g h i 32 Solutions 169 ad be cf a b c 20 d e f d e f 7 g h i g h i 2 3 0 21 Since 1 3 4 1 0 the matrix is invertible 1 2 1 5 0 1 22 Since 1 3 2 0 the matrix is not invertible 0 5 3 2 0 0 8 1 7 5 0 23 Since 0 the matr1x 1s not 1nve1t1ble 3 8 6 0 0 7 5 4 7 3 24 Since 6 0 5 11 0 the columns of the matrix form a linearly independent set 7 2 6 7 8 7 25 Since 4 5 0 1 0 the columns of the matrix form a linearly independent set 6 7 5 3 2 2 0 5 6 1 26 Since 6 0 3 0 0 the columns of the matr1x form a lmearly dependent set 4 7 0 3 27 a True See Theorem 3 b True See the paragraph following Example 2 c True See the paragraph following Theorem 4 d False See the warning following Example 5 28 a True See Theorem 3 b False See the paragraphs following Example 2 c False See Example 3 d False See Theorem 5 29 By Theorem 6 det35 detB5 25 32 30 Suppose the two rows of a square matrixA are equal By swapping these two rows the matrixA is not changed so its determinant should not change But since swapping rows changes the sign of the determinant detA 7 detA This is only possible if detA 0 The same may be proven true for columns by applying the above result to AT and using Theorem 5 170 40 6 One may compute using Theorem 2 that detA 3 and detB 8 while AB 17 One may compute that detA 0 and det B 72 while AB 2 detAB CHAPTER3 Determinants By Theorem 6 detAdetA 1 det 1 so detA l 1detA By factoring an r out of each of the n rows detrA rndetA By Theorem 6 detAB detAdet B det BdetA det BA By Theorem 6 and Exercise 31 detPAP 1detPdetAdetP 1 detPdetP 1detA 1 det P detP detA1detA detA By Theorem 6 and Theorem 5 detUTU detUTxdetU detU2 Since UTU 1 detUTU det 1 so detU2 1 Thus det U i1 By Theorem 6 detA4 detA4 Since detA4 0 then detA4 0 Thus detA 0 andA is not invertible by Theorem 4 0 Thus 4 detAB 24 3 x 8 detAdet B 6 0 0 Thus detAB 0 0 x 72 detAdet B a By Theorem 6 detAB detAdet B 4 X 73 712 b By Exercise 32 det 5A 53 detA 125gtlt 4 500 c By Theorem 5 detBT detB 3 d By Exercise 31 detA l 1detA 1 4 e By Theorem 6 detA3 detA3 43 64 a By Theorem 6 detAB detAdet B 71 X 2 72 b By Theorem 6 detB5 detB5 25 32 c By Exercise 32 det 2A 24 detA 16gtlt 1 16 d By Theorems 5 and 6 detATA detATdetA detAdetA 1gtlt 1 1 e By Theorem 6 and Exercise 31 detB lAB detB 1detAdetB 1 detBdetAdetB detA 1 detA a edicb f adedibcicf adibc edicf detB detC 1a 1d det A B detA detBifand only ifa d 0 b 1a1d cb1adad cbdetAaddetBso 33 Solutions 171 43 Compute detA by using a cofactor expansion down the third column detA u1 v1detA13 u2 v2 detA23 113 v3detA33 uldetA13 uzdetA23 u3detA33 vldetA13 vzdetA23 v3detA33 detB detC 44 By Theorem 5 detAE detAET Since AET ETAT detAE detETAT Now ET is itself an elementary matrix so by the proof of Theorem 3 det ETAT detET detAT Thus it is true that detAE detET detAT and by applying Theorem 5 detAE det EdetA 45 M Answers will vary but will show that detAT A always equals 0 while detAAT should seldom be zero To see why AT A should not be invertible and thus detAT A 0 letA be a matrix with more columns than rows Then the columns of A must be linearly dependent so the equationAx 0 must have a nontrivial solution x Thus ATAX AT Ax AT 0 0 and the equation ATAX 0 has a nontrivial solution Since AT A is a square matrix the Invertible Matrix Theorem now says that AT A is not invertible Notice that the same argument will not work in general for AAT since AT has more rows than columns so its columns are not automatically linearly dependent 46 M One may compute for this matrix that detA 1 and condA 23683 Note that this is the 2 condition number which is used in Section 23 Since detA 0 it is invertible and 19 14 0 7 549 401 2 196 267 195 1 95 278 203 1 99 AT1 The determinant is very sensitive to scaling as det10A 104 detA 10000 and det01A 014detA 00001 The condition number is not changed at all by scaling cond 10A cond01A condA 2 23683 When A 14 detA1 and condA 1 As before the determinant is sensitive to scaling det10A 104detA 10 000 and det01A 014 detA 00001 Yet the condition number is not changed by scaling cond10A cond01A condA 1 33 SOLUTIONS Notes This section features several independent topics from which to choose The geometric interpretation of the determinant Theorem 10 provides the key to changes of variables in multiple integrals Students of economics and engineering are likely to need Cramer s Rule in later courses Exercises 1710 concern Cramer s Rule exercises 11718 deal with the adjugate and exercises 1932 cover the geometric interpretation of the determinant In particular Exercise 25 examines students understanding of linear independence and requires a careful explanation which is discussed in the Study Guide The Study Guide also contains a heuristic proof of Theorem 9 for 2 X 2 matrices Equot N 43 A El39 CHAPTER3 Determinants 5 7 3 The system is equivalent to Ax b where A 2 4 and b We compute 3 7 5 3 4111 4A2b2 1detA6detA1b5detA2b L detA1b 5 detA 6 x detA2bl 2 detA 639 4 1 6 The system is equivalent to Ax b where A 5 2 and b 7 We compute 6 1 4 6 A1b7 2A2b5 7detA3detA1b5detA2b 2 detA1b 5 detA2b 2 2 detA 3 detA 339 3 2 7 The system is equivalent to Ax b where A 5 6 and b 5 We compute 7 2 3 7 21115 6A2b5 5detA8detA1b32detA2b20 detA1b 32 detA 8 detA2b 20 5 detA 8 239 4 x2 5 3 9 The system is equivalent to Ax b where A 3 I and b 5 We compute 9 3 5 9 A1b 5 l A2 b 3 5 detA 4 detA1b 6 detA2 b 2 x detA1b 6 3 1 x2 detA2b 2 1 detA 4 2 detA 4 239 2 1 0 7 The system is equivalent to Ax b where A 3 0 l and b 8 We compute 0 1 2 3 7 1 0 2 7 0 2 1 7 A1b 8 0 1 A2b 3 8 1 A3b 3 0 8 3 l 2 0 3 2 0 1 3 detA 4detA1b 6detA2b 16 detA3b 14 detA1b 6 3 x detA2b 16 detA 4 2 2 detA 4 detA3b 14 7 detA 4 239 4 33 Solutions 173 2 1 1 4 6 The system is equivalent to AX b where A 1 0 2 and b 2 We compute 3 1 3 2 4 1 1 2 4 1 2 1 4 A1b 2 0 2 A2b l 2 2 A3b l 0 2 2 1 3 3 2 3 3 l 2 detA 4 detA1b 16detA2b 52 detA3b 4 detA1b 16 4 x detA2b 52 13 x detA3b 4 detA 4 2 detA 4 a 3 1 detA 4 6 4 5 7 The system is equivalent to Ax b where A g 2 j and b 2 We compute S 5 4 6 5 21112 2SA2b 2detA1b1038detA2b 12s 45 Since detA 1232 36 12s2 3 0 for 3 33 the system will have a unique solution when 3 iJ3 For such a system the solution will be detA1b 1038 SS 4 x detA2b 12s 45 4s 15 detA 12s2 3 6s2 3 2 detA 12s2 3 4s2 3 39 3 5 3 8 The system is equivalent to Ax b where A g 5 and b 2 We compute s 3 5 3 3 4112 5SA2b 2detA1b15310detA2b65 27 Since detA 1532 45 15s2 3 0 for all values of s the system will have a unique solution for all values of s For such a system the solution will be x detA1b 15510 3s2 x detA2b 63 27 25 9 1 detA 15s2 3 3s2 3 2 detA 15s2 3 5s2 3 39 3 2s 1 9 The system 1s equ1valent to Ax b where A 3 6 and b 4 We compute s 1 2s 3 1 A1b A2b detA1b2s detA2b 4s3 4 6s 3 4 Since detA 632 63 6ss 1 0 for s 0 71 the system will have a unique solution when 3 0 71 For such a system the solution will be x detA1b 2s 1 x detA2b 4s3 I 1 detA 6ss 1 3s 1 2 detA 6ss 1 2 l 1 10 The system is equivalent to Ax b where A 3S 6 J and b 2 We compute s s 1 1 2411 2 6S A2b detA1b 6s 2 detA2b s 174 CHAPTER3 Determinants Since detA 1232 3s 3s4s 1 0 for s 014 the system will have a unique solution when 3 0 1 4 For such a system the solution will be detA1b 63 2 x detA2b s 1 detA 3s4s 1 2 detA 3s4s 1 34s 1 11 Since detA 3 and the cofactors of the given matrix are 0 0 3 0 3 0 0111 10 Cu 1 1 3 C131 13 2 1 0 1 0 2 C21 1 11 C222 1 11 C23 1 12 2 1 0 1 0 2 C31 0 00 C323 0 3 C333 06 0 1 0 0 13 0 ade 3 1 3 and1 1 1 ade 1 13 1 detA 3 2 6 1 23 2 12 Since detA 5 and the cofactors of the given matrix are C 2 1 1 C 2 10 C 2 2 2 11 1 0 a 12 0 0 a 13 0 1 C 1 3 3 C 1 3 0 C 1 1 L 21 10quot 22 0 0quot 23 01 C 1 3 7 C 1 3 5 C 1 1 4 31 2 1 5 32 2 1 5 33 2 2 5 1 3 7 15 35 75 ade 0 0 5 and A4 1 ade 0 0 1 detA 2 1 4 25 15 45 13 Since detA 6 and the cofactors of the given matrix are C 0 1 L C 1 1L C 1 0L 11 1 1 12 2 1 13 2 1 C 5 4 1 C 3 4 5 C 3 5 7 21 1 1 a 22 2 1 a 23 2 1 a C 5 4 5 C 3 4 1 C 3 5 5 31 0 1 5 32 1 1 5 33 1 0 5 1 1 5 16 16 56 ade 1 5 landA lz 1 ade 16 56 16 detA 1 7 5 16 76 56 14 Since detA 71 and the cofactors of the given matrix are 2 1 0 1 0 2 C 3 4 25 C12 2 4 22 C13 2 3 24 6 7 3 7 3 6 C213 3 3 C22 2 4 2 C232 33 C 6 7 8 C 3 7 3 C 3 6 6 31 2 1 a 32 0 1 a 33 0 2 a 5 3 8 1 5 3 8 ade 2 2 3 and A4 ade 2 2 3 detA 4 3 6 4 3 6 15 Since detA 6 and the cofactors of the given matrix are C102 C 102 C 11L 11 3 2quot 2 2quot 13 2 3 0 0 3 0 3 0 C 3 20 C222 2 26 C23 2 3 9 C 0 0 0 C 0 0 0 C 3 0 3 31 1 0 31 1 0 331 1 2 0 0 13 0 0 ade 2 6 0 and A4 1 ade 13 1 0 detA 1 9 3 16 32 12 16 Since detA 79 and the cofactors of the given matrix are 3 1 0 1 0 3 C11 0 3 9 C120 30 C130 00 C 2 4 6 C 1 4 3 C 1 2 0 21 0 3 5 22 0 3 5 23 0 0 5 C 2 4 14 C 1 4 1 C 1 2 3 3l3 1 5 32 0 1 5 33 0 3 5 9 6 14 1 23 149 ade 0 3 1 and A4 1 ade 0 13 19 detA 0 0 3 0 0 13 b a 17 Let A a d j Then the cofactors ofA are C11 ldl d C12 c c 33 Solutions 175 d b C21 b b and C22 lal 1 Thus ade Since detA adi be Theorem 8 gives that c a d b A71 L de This result is identical to that of Theorem 4 in Section 22 a detAa Zad bc c gt0 N G N b N A CHAPTER3 Determinants Each cofactor of A is an integer since it is a sum of products of entries inA Hence all entries in adj A will be integers Since detA l the inverse formula in Theorem 8 shows that all the entries in A 1 will be integers 5 6 The parallelogram is determined by the columns of A 2 4 so the area of the parallelogram is ldetAl w 8 1 4 The parallelogram is determined by the columns of A 3 5 so the area of the parallelogram is ldetAl Hi 7 First translate one vertex to the origin For example subtract 1 0 from each vertex to get a new parallelogram with vertices 0 0l 52 74 and 3 1 This parallelogram has the same area as the original and is determined by the columns of A so the area of the parallelogram is 1 2 5 4 ldetAll714l 14 First translate one vertex to the origin For example subtract 0 72 from each vertex to get a new parallelogram with vertices 0 06 l73 3 and 3 4 This parallelogram has the same area as the original and is determined by the columns of A 1 i3 so the area of the parallelogram is ldetAl l2ll 21 1 1 7 The parallelepiped is determined by the columns of A 0 2 l so the volume of the 2 4 0 parallelepiped is ldetAl l22l 22 1 2 1 The parallelepiped is determined by the columns of A 4 5 2 so the volume of the 0 2 1 parallelepiped is ldetAl lilSl 15 The Invertible lVIatrix Theorem says that a 3 X 3 matrixA is not invertible if and only if its columns are linearly dependent This will happen if and only if one of the columns is a linear combination of the others that is if one of the vectors is in the plane spanned by the other two vectors This is equivalent to the condition that the parallelepiped determined by the three vectors has zero volume which is in turn equivalent to the condition that detA 0 By de nition p S is the set of all vectors of the form p v where v is in S Applying T to a typical vector in p S we have Tp v Tp Tv This vector is in the set denoted by Tp TS This proves that Tmaps the set p S into the set Tp TS Conversely any vector in T p T S has the form T p T v for some v in S This vector may be written as T p v This shows that every vector in T p T S is the image under T of some point p v in p S 27 4 Since the parallelogram S is determined by the columns of 33 Solutions 177 2 5 the area ofSis 2 Since the parallelogram S is determined by the columns of 3 2 2 det 3 5 ldetAHarea ofS 6 4 24 Alternatively one may compute the vectors that determine the image namely the columns of mw a rma The determinant of this matrix is 724 so the area of the image is 24 6 2 l 4 l 4 The matrixA has detA 3 2 6 By Theorem 10 the area of TS is 0 the area ofS is 7 1 4 0 7 2 det 7 I l 4 4 The matrixA has detA 1 1 5 By Theorem 10 the area ofTS is ldetAHarea ofS 5 4 20 Alternatively one may compute the vectors that determine the image namely the columns of Am WE ill 3 3H3 ii The determinant of this matrix is 20 so the area of the image is 20 The area of the triangle will be one half of the area of the parallelogram determined by v1 and v2 By Theorem 9 the area of the triangle will be l2ldetAl where A v1 v2 Translate R to a new triangle of equal area by subtracting g y3 from each vertex The new triangle has vertices 0 0 x1 gy1 y3 and x2 gy2 y3 By Exercise 29 the area ofthe triangle will be napalm 2 yl y3 y2 y3 Now consider using row operations and a cofactor expansion to compute the determinant in the formula x1 y1 1 x1x3 y1y3 0 xx yy det x2 y2 lzdet x2x3 y2y3 0det1 3 1 3 x2 yz y3 x3 3 3 1 x3 y3 1 By Theorem 5 detx1 y1y3detx1 x3 x2 x2 y2y3 y1y3 y2y3 So the above observation allows us to state that the area of the triangle will be xrxs x x 1 xi yl 1 det 2 3 det x2 y2 l y1y3 y2y3 2 yg 1 l 2 178 CHAPTER3 Determinants 1 2 2 2 31 a To show that T S is bounded by the ellipsoid with equation 2 x 1 let u M2 and let a c 3 x1 x x2 Au Then u1x1a u2 x2b and M3 x3c anduliesinsideSor u12u22u3231if x3 x12 x2 x2 2 3 and only 1fx lies ms1de TS or yb 2C ZS 1 b By the generalization of Theorem 10 volume of ellipsoid volume of T 5 detA volume of S abcgm 32 a A linear transformation T that maps S onto 539 will map e1 to v1 e2 to v2 and e3 to v3 that is T el v1 T e2 v2 and T e3 v3 The standard matrix for this transformation will be A Te1 Te2 Te3 v1 v2 v3 1 The area ofthe base ofS is 1211 12 so the volume ofS is 13121 16 By part a T S S 39 so the generalization of Theorem 10 gives that the volume of S39 is detAvolume of S 16detA P 33 M Answers will vary In MATLAB entries m3 7 i11vA are approximately 10 15 or smaller 34 M Answers will vary as will the commands which produce the second entry of x For example the MATLAB command is X2 det A z l b A z 34 det A while the Mathematica command is X2 Det Transpose A 1 bTranspose A 3 Transpose A 4 Det A 35 M MATLAB Student Version 40 uses 57771 ops for invA and 14269045 ops for the inverse formula The inv A command requires only about 04 of the operations for the inverse formula Chapter 3 SUPPLEMENTARY EXERCISES 1 a True The columns ofA are linearly dependent b True See Exercise 30 in Section 32 0 False See Theorem 3c in this case det 5A 53det A 2 0 1 0 3 0 d False Cons1der A B and A B 0 1 0 3 0 4 e False By Theorem 6 det A3 23 f False See Theorem 3b g True See Theorem 3c h True See Theorem 3a i False See Theorem 5 j False See Theorem 3c this statement is false for n X n invertible matrices with n an even integer k True See Theorems 6 and 5 det ATA det A2 4 l Chapter 3 Supplementary Exercises 179 1 False The coefficient matrix must be invertible m False The area ofthe triangle is 5 n True See Theorem 6 detA3 det A3 0 False See Exercise 31 in Section 32 p True See Theorem 6 12 13 14 12 13 14 15 16 17 3 3 3 0 18 19 20 6 6 6 l a bc l a bc 1 a bc 1 b 10 0 b a a b b ac a 0 l 1 0 l c ab 0 6 11 11 0 0 l 1 a b c a b c a b c ax bx cxx x xxyl 1 10 ay by cy y y y l l l 9 1 9 9 9 9 9 9 2 9 0 9 9 2 4 0 5 4005 0140 5 012939 9 3 9 0 9 0 3 9 0 6 0 7 6 0 7 0 6 0 0 7 0 4 5 123 6 7 123212 4 8 8 8 5 4 8 8 5 0 1 0 0 0 4 8 5 6 8 8 8 7 D 6 8 8 7 12 6 8 7 12 3 4 5 12 3 2 12 0 8 3 0 6 7 0 8 8 3 0 0 3 0 0 2 0 0 0 8 2 0 0 Expand along the first row to obtain 1 x y 1 x1 y1 y1 1 x1 7 x1 y1 1 x y 0 Th1s is an equatlon ofthe formwcby070 1 x y x2 3 2 1 3 2 x2 2 2 and since the points x1y1 and x2 y2 are distinct at least one of a and b is not zero Thus the equation is the equation of a line The points x1 y1 and x2 y2 are on the line because when the coordinates of one of the points are substituted for x and y two rows of the matrix are equal and so the determinant 1s zero 180 CHAPTER3 Determinants 8 Expand along the first row to obtain 1 x y 1 y1 1 y 0 0 m xlyl 1m lxry11 01m rewrittenas rmcl yl mxy0 or y y1mx x1 1 1rmc1 y1 xm y1 0 This equation may be l a a2 l a a2 1 a a2 9 detTl b b2 0 b a bZ a2 0 b a b aba l c 02 0 0 11 02 112 0 0 C aXC l39a l a a2 l a a2 b ac a 0 l ba b ac a 0 l ba b ac ac b 0 1 61 0 0 c b 10 Expanding along the first row will show that f t det V co 01t cztz 0313 By Exercise 9 1 xi xf cg1 x2 x22 x2JqJeJqX3XZ 0 1 Ia x32 since x1 x2 and g are distinct Thusft is a cubic polynomial The points x10 9620 and x30 are on the graph off since when any of x1 x2 or x3 are substituted for t the matrix has two equal rows and thus its determinant which is f 1 is zero Thus f x 0 for i 1 2 3 11 To tell if a quadrilateral determined by four points is a parallelogram rst translate one of the vertices to the origin If we label the vertices of this new quadrilateral as 0 v1 v2 and v3 then they will be the vertices of a parallelogram if one of v1 v2 or v3 is the sum of the other two In this example subtract 1 4 from each vertex to get a new parallelogram with vertices 0 0 0 v1 2 1 v2 2 5 and v3 44 Since v2 v3 v1 the quadrilateral is a parallelogram as stated The translated parallelogram has the same area as the original and is determined by the columns of 2 4 A v1 v3 1 4 so the area ofthe parallelogram is ldetAl 17121 12 12 A 2 X 2 matrixA is invertible if and only if the parallelogram determined by the columns of A has nonzero area 1 d tA A A 1A I By the Invertible Matrix Theorem adj A is invertible and e 13 By Theorem 8 adj AA 1 dA 1 A a1 detA A O 14 21 Consider the matrix A O I j where l S k S n and O is an appropriately sized zero matrix We k will show that det At det A for all 1 S k S n by mathematical induction P O b Chapter 3 Supplementary Exercises 181 First let k 1 Expand along the last row to obtain A O det A1 det 1ltquot1ltquot1 A 1 det A det A O 1 Now let 1 lt k S n and assume that det Ak1 det A Expand along the last row of Af to obtain A O nknr det Ar det O I l l det Ak1 det Ak1 det A Thus we have proven the result k and the determinant of the matrix in question is detA I 0 Consider the matrix A Ck D where l S k S n C k is an n X k matrix and O is an appropriately k sized zero matrix We will show that detAf detD for all 1 S k S n by mathematical induction First let k 1 Expand along the rst row to obtain 1 O 11 detA1 det l lAdetD detD C1 D Now let 1 lt k S n and assume that detAk1 detD Expand along the rst row of Af to obtain 1 O detA6 det Ck D l11 A l detAk1 detAJH det D Thus we have proven the result and the k determinant of the matrix in question is det D By combining parts a and b we have shown that detA OdetA ODdetI ODdetAdetD C D O I C D From this result and Theorem 5 we have A B A B T AT det det det O D O D BT Compute the right side of the equation I O A B A B X 1 0 Y Ql XB Y Set this equal to the left side of the equation A B A B C D Ql XB Y SinceM C andA is invertible X CA 1 SinceXB Y D Y D XB D CA IB Thus by Exercise l4c A B I O A B det det det C D CA 1 1 0 D CA IB detAdetD CA IB 1detATXdetDT detAdetD Jsothat QlC XBYD From part a A B 1 71 det C D detAdet D CA B detAD CA B detAD ACA IB detAD CAA IB det AD CB 182 CHAPTER3 Determinants 16 a Doing the given operations does not change the determinant of A since the given operations are all p A H row replacement operations The resulting matrix is Ta b 7ab 0 0 0 a b 7ab 0 0 0 a b 0 b b b a b Since column replacement operations are equivalent to row operations on AT and detAT detA the given operations do not change the determinant of the matrix The resulting matrix is Ta b 0 0 0 0 a b 0 0 0 0 a b 0 b 2b 3b an lb c Since the preceding matrix is a triangular matrix with the same determinant as A detA a7 b 1an 1b First consider the case n 2 In this case 7b b b b detBa aa bdetC ab b2 a b a so detA detB detC aa b ab 132 a2 b2 a7 ba b a7 bHa 2 1b and the formula holds for n 2 Now assume that the formula holds for all k 7 l X k 7 1 matrices and letA B and C be kgtlt k matrices By a cofactor expansion along the first column a b b a detB a7 b a ba bHa k 2b a bHa k 2b b b 1 since the matrix in the above formula is a k 7 l X k 7 1 matrix We can perform a series of row operations on C to zero out below the rst pivot and produce the following matrix whose determinant is det C b b b 0 a b 0 0 0 a b Since this is a triangular matrix we have found that detC ba by 1 Thus detA detB detC a bHa k 2b ba bH a bHa k 1b which is what was to be shown Thus the formula has been proven by mathematical induction M Since the first matrix has a 3 b 8 and n 4 its determinant is 3 8 3 4 18 53 3 24 12527 3375 Since the second matrix has a 7 8 b 7 3 and n 7 5 its determinant is 8 3H8 5 13 548 12 62520 12 500 19 M We nd that Our conjecture then is that 1 1 1 1 1 2 2 2 1 2 3 3 Chapter 3 Supplementary Exercises 183 11111 111 12 2 2 2 222 112 3 3 31 23 3 12 3 4 4 23 4 1 2 3 4 5 1 2 3 71 To show this consider using row replacement operations to zero out below the rst pivot The resulting matrix is 1 1 1 1 0 0 1 1 2 012 n l Now use row replacement operations to zero out below the second pivot and so on The nal matrix which results from this process is 71 0 1 1 001 1 1 7000 1 1 1 1 which is an upper triangular matrix with determinant 1 1 1 1 1 3 3 3 M We nd that 11111 111 13333 333 181366654 366 13699 369 136912 1 3 623 184 CHAPTER3 Determinants To show this consider using row replacement operations to zero out below the first pivot The resulting matrix is 1 1 1 1 0 2 2 0 2 5 5 0 2 5 3n l l Now use row replacement operations to zero out below the second pivot The matrix which results from this process is 1 1 1 l 1 1 1 0 2 2 2 2 2 2 0 0 3 3 3 3 3 0 0 3 6 6 6 6 0 0 3 6 9 9 9 70 0 3 6 9 12 3n 2 This matrix has the same determinant as the original matrix and is recognizable as a block matrix of the orm A B lo D1 where 3 3 3 3 3 1 1 1 1 1 3 6 6 6 6 l 2 2 2 A 1andD 3 6 9 9 9 31 2 3 3 3 0 2 I I 3 6 9 12 3n 2 1 2 3 4 n 2 A B As in Exercise l4c the determinant of the matrix 0 D is detAdet D 2 det D Since D is an n 7 2 X n 7 2 matrix 1 1 1 1 1 1 2 2 2 2 detD3H1 2 3 3 3 3H13 2 1 2 3 4 n 2 A B by Exercise 19 Thus the determinant of the matrix 0 D is 2detD 23 Matrix Algebra 21 SOLUTIONS Notes The de nition here of a matrix product AB gives the proper view of AB for nearly all matrix calculations The dual fact about the rows of A and the rows of AB is seldom needed mainly because vectors here are usually written as columns I assign Exercise 13 and most of Exercises 17 22 to reinforce the de nition of AB Exercises 23 and 24 are used in the proof of the Invertible Matrix Theorem in Section 23 Exercises 23 25 are mentioned in a footnote in Section 22 A class discussion of the solutions of Exercises 23 25 can provide a transition to Section 22 Or these exercises could be assigned after starting Section 22 Exercises 27 and 28 are optional but they are mentioned in Example 4 of Section 24 Outer products also appear in Exercises 3 l 34 of Section 46 and in the spectral decomposition of a symmetric matrix in Section 7 l Exercises 29 33 provide good training for mathematics majors 2 0 l 4 0 2 1 2A 2 2 Next use B 2A B 2A 4 5 2 8 10 4 7 5 l 4 0 2 3 5 3 l 4 3 8 10 4 7 6 7 The product AC is not de ned because the number of columns of A does not match the number of rows 1 2 3 5 l32 l l524 l 13 of C CD For mental computatlon the 2 l l 4 23l l 25l4 7 6 rowcolumn rule is probably easier to use than the de nition 2 0 l 7 5 l 2l4 0 10 l2 16 10 l 2 A 23 2 4 5 2 l 4 3 42 5 8 2 6 6 l3 4 The expression 3C E is not de ned because 3C has 2 columns and E has only 1 column CB 1 2 7 5 1 1721 1 52 4 112 3 9 13 5 2 111 4 3 2711 2 51 4 211 3 13 6 4 The product EB is not de ned because the number of columns of E does not match the number of rows of R 83 84 CHAPTER2 MatIiXAlgebra 3 0 4 4 3 4 04 1 1 1 3 312 24 0 3 5 2 0 5 3 2 5 5 4 1 12 3 312A312A35 or 2 15 6 3 0 4 1 3440 3 10 12 3 312A 0 3 5 2 03 5 03 2 15 6 9 1 3 5 0 0 4 1 3 4 A 513 8 7 6 0 5 0 8 2 6 4 1 8 0 0 5 4 1 3 9 1 3 45 5 15 513A5I3A5A5 8 7 6 40 35 30 or 5 0 0 9 1 3 513A 0 5 0 8 7 6 0 0 5 4 1 8 5900 5 100 5300 45 5 15 05 80 0570 05 60 45 35 30 005 4 0051 0058 20 5 40 1 2 3 7 1 2 2 4 5 a Ab1 5 42 7 Ab2 5 4 I 6 2 3 12 2 3 7 7 4 21152424151 21152 7 6 12 7 1 2 132 2 1 221 7 4 3 21 b 5 4 534 2 5241 7 6 2 2 3 23 3 2 2 2 31 12 7 4 2 1 0 4 2 3 14 6 a Ab1 3 0 M 3 Ab2 3 0 I 9 3 5 13 3 5 4 0 14 2115242411 21152 3 9 13 4 4 2 411 22 413 2 1 0 14 1 3 b 3 0 3102 330 1 3 9 3152 335 1 13 4 9 gt0 N p A DJ p A A Use the de nition of AB written in reverse order Abl 1 1 1 Abp Ab1 1 1 1 By de nition UQ Uq1 1 1 1 q Uql 21 Solutions 85 SinceA has 3 columns B must match with 3 rows Otherwise AB is unde ned Since AB has 7 columns so does B Thus B is 3gtlt7 The number of rows of B matches the number of rows of BC so B has 3 rows 10 5k 23 15 AB254 523 525 3 13 k 915k k 316 3k15k39 ThenAB BA ifand only if710 5k 15 and 79 6 7 3k which happens ifand only ifk 5 a ill 1 3 at 21 ll 111 4 while BA 3 2 1 1 1 2 0 0 2 5 AD 1 2 3 0 3 0 2 6 15 1 4 5 0 0 5 2 12 25 2 0 0 1 1 1 2 2 2 DA 0 3 0 1 2 3 3 6 9 0 0 5 1 4 5 5 20 25 Rightmultiplication that is multiplication on the right by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D Leftmultiplication by D multiplies each raw of A by the corresponding diagonal entry of D To make AB BA one can take B to be a multiple of 13 For instance ifB 413 thenAB and BA are both the same as 4A Consider B b1 b2 To make AB 0 one needs Abl 0 andAbz 0 By inspection ofA a suitable 2 2 2 6 his 1 or any multiple of 1 Example B 1 3 hp Thus QrpQRwhenRr1 rp Q 1 1 1 Uq4 From Example 6 of Section 18 the vector Uql lists the total costs material labor and overhead con esponding to the amounts of products B and C speci ed in the vector ql That is the first column of UQ lists the total costs for materials labor and overhead used to manufacture products B and C during the first quarter of the year Columns 2 3 and 4 of UQ list the total amounts spent to manufacture B and C during the 2nd 3 and 439h quarters respectively 21 False See the de nition of AB False The roles of A and B should be reversed in the second half of the statement See the box after Example 3 True See Theorem 2b read right to left True See Theorem 3b read right to left False The phrase in the same order should be in the reverse order See the box after Theorem 3 Uquot P DQ False AB must be a 3gtlt3 matrix but the formula for AB implies that it is 3gtlt 1 The plus signs should be just spaces between columns This is a common mistake F 9 True See the box after Example 6 False The lefttoright order of B and C cannot be changed in general P 86 CHAPTER2 MatrixAlgebra d False See Theorem 3d e True This general statement follows from Theorem 3b 1 2 17 Since 6 9 3 AB Ab1 Ab2 Ab3 the rst column ofB satis es the equation 1 l 2 l l 0 7 7 Ax Row reduction A Ab1N N So b1 Similarly 6 2 5 6 0 l 4 4 l 2 2 l 0 8 8 A Ab2N N and b2 2 5 9 0 l 5 5 Note An alternative solution of Exercise 17 is to row reduce A Abl Abz with one sequence of row operations This observation can prepare the way for the inversion algorithm in Section 22 H 8 The rst two columns of AB areAb1 andAbz They are equal since b1 and b2 are equal H 9 A solution is in the text Write B b1 b2 b3 By de nition the third column of AB is Ab3 By hypothesis b3 b1 b2 So Ab3 Ab1 b2 Abl Ab2 by a property of matrixvector multiplication Thus the third column of AB is the sum of the rst two columns of AB 20 The second column ofAB is also all zeros becauseAbz A0 0 21 Let bp be the last column of B By hypothesis the last column ofAB is zero Thus Abp 0 However bp is not the zero vector because B has no column of zeros Thus the equationAbp 0 is a linear dependence relation among the columns of A and so the columns of A are linearly dependent Note The text answer for Exercise 21 is quotThe columns of A are linearly dependent Why The Study Guide supplies the argument above in case a student needs help 22 If the columns of B are linearly dependent then there exists a nonzero vector x such thath 0 From this ABx A0 and ABX 0 by associativity Since x is nonzero the columns ofAB must be linearly dependent N b Ifx satis es Ax 0 then CAx C0 0 and so Inx 0 and x 0 This shows that the equationAx 0 has no free variables So every variable is a basic variable and every column of A is a pivot column A variation of this argument could be made using linear independence and Exercise 30 in Section 17 Since each pivot is in a different row A must have at least as many rows as columns 24 Take any h in R By hypothesisADb Imb b Rewrite this equation as ADb b Thus the vector x Db satis es Ax b This proves that the equationAx b has a solution for each h in R By Theorem 4 in Section 14 A has a pivot position in each row Since each pivot is in a different columnA must have at least as many columns as rows 25 By Exercise 23 the equation CA 1 implies that number of rows inA 3 number of columns that is m 3 n By Exercise 24 the equation AD 1m implies that number of rows inA number of columns that is m E n Thus m 11 To prove the second statement observe that DAC DAC InC C and also DAC DAC DIm D Thus C D A shorter calculation is C 1C DAC DAC D1 D 26 Write 3 e1 e2 e3 and D d1 d2 d3 By de nition ofAD the equation AD 13 is equivalent lto the three equations Adl e1 Adz e2 andAd3 e3 Each of these equations has at least one solution because the columns ofA span R3 See Theorem 4 in Section 14 Select one solution of each equation and use them for the columns ofD Then AD 13 21 Solutions 87 27 The product uTv is a 1gtlt1 matrix which usually is identi ed with a real number and is written without the matrix brackets 11 2 uTv 2 3 4 b 2a3b 4c vTua b c 3 2a3b 4c c 4 2 2a 2b 20 uvT 3 a b 0 3a 3b 3c 4 4a 4b 4c 1 2a 3a 4a vuT b 2 3 4 2b 3b 4b 0 20 30 4c 28 Since the inner product uTv is a real number it equals its transpose That is uTv uTvT vTuTT vTu by Theorem 3d regarding the transpose of a product of matrices and by Theorem 3a The outer product uvT is an n Xn matrix By Theorem 3 uvTT VTTuT vu 29 The 139 jentry of AB 0 equals the 139 jentry ofAB AC because 2 1 ka ch Z atij Z awe k1 k1 151 The 139 jentry of B CA equals the 139 jentry of BA CA because 2 blk 521an Z bzkacy Z akac7 k1 k1 151 30 The 139 j entries of rAB rAB andArB are all equal because 2 azkbky Z quot121th Z azkrby Ir1 k1 k1 31 Use the de nition of the product ImA and the fact that Imx x for x in R ImA Ima1 an1ma1 Iman 211 an A 32 Let e and a denote the jth columns of In andA respectively By de nition the jth column of AIquot isAej which is simply 21 because e has 1 in the jth position and zeros elsewhere Thus corresponding columns ofAIn andA are equal Hence A1quot A 33 The 139 jentry of ABT is the j ientry ofAB which is a bll ma b 1n m The entries in row i of BT are b1 bm because they come from column 139 of B Likewise the entries in columnj ofAT are 11 aw because they come from rowj ofA Thus the ijentry in ETAT is a bll a an as above 34 Use Theorem 3d treating x as an ngtltl matrix Al3xT XTABT XTBTA 7 35 M The answer here depends on the choice of matrix program For MATLAB use the help command to read about zeros ones eye and diag For other programs see the appendices in the Study Guide The TI calculators have fewer single commands that produce special matrices 88 CHAPTER2 MatrixAlgebra 36 M The answer depends on the choice of matrix program In MATLAB the command rand 6 4 creates a 6gtlt4 matrix with random entries uniformly distributed between 0 and l The command round 19 rand 6 4 5 creates a random 6gtlt4 matrix with integer entries between 79 and 9 The same result is produced by the command randomint in the Laydata Toolbox on text website For other matrix programs see the appendices in the Study Guide 43 l M A IA 7 7 A2 7 0 for all 4gtlt4 matrices However A BA 715 7A2 732 is the zero matrix only in the special cases when AB BA In general A BA iB AA 73 BA 715 AA 7AB BA 733 DJ on M The equality ABT ATBT is very likely to be false for 4gtlt4 matrices selected at random 39 M The matrix S shifts the entries in a vector 11 b c d e to yield b c d e 0 The entries in S2 result from applying S to the columns of S and similarly for S 3 and so on This explains the patterns of entries in the powers of S 00100 00010 00001 00010 0000100000 520 0 0 01S30 0 0 0 0S40 0 0 0 0 00000 00000 00000 00000 00000 00000 S5 is the 5gtlt5 zero matrix S6 is also the 5gtlt5 zero matrix 3318 3346 3336 333337 333330 333333 40 M A5 3346 3323 33311410 333330 333336 333334 3336 3331 3333 333333 333334 333333 The entries in A20 all agree with 3333333333 to 9 or 10 decimal places The entries in A30 all agree with 33333333333333 to at least 14 decimal places The matrices appear to approach the matrix 1 3 1 3 1 3 l 3 13 13 Further exploration of this behavior appears in Sections 49 and 52 1 3 1 3 1 3 Note The MATLAB box in the Study Guide introduces basic matrix notation and operations including the commands that create special matrices needed in Exercises 35 36 and elsewhere The Study Guide appendices treat the corresponding information for the other matrix programs 22 SOLUTIONS Notes The text includes the matrix inversion algorithm at the end of the section because this topic is popular Students like it because it is a simple mechanical procedure However I no longer cover it in my classes because technology is readily available to invert a matrix whenever needed and class time is better spent on more useful topics such as partitioned matrices The nal subsection is independent of the inversion algorithm and is needed for Exercises 35 and 36 Key Exercises 8 11724 35 Actually Exercise 8 is only helpful for some exercises in this section Section 23 has a stronger result Exercises 23 and 24 are used in the proof of the Invertible Matrix Theorem IMT in Section 23 along with Exercises 23 and 24 in Section 2 l I recommend letting students work on two or more of these four exercises before proceeding to Section 23 In this way students participate in the 22 Solutions 89 proof of the IMT rather than simply watch an instructor carry out the proof Also this activity will help students understand why the theorem is true 186 114 6 2 3 395 4 32 30 5 8 52 4 2 3 2 1 1 4 2 2 1 39 7 4 12 14 7 3 72 32 8 5 1 1 5 5 1 5 5 1 1 7 5 40 35 7 8 5 7 8 14 16 43 4 1 1 841 84 2 1 01 7 8 24 28 7 3 4 7 3 74 34 8 6 2 5 The system 1s equlvalent to Ax b whereA 5 4 and b 1 and the solutlon 1s 71 2 3 2 7 XA b Thusx17andxz79 5 2 4 1 9 I 8 5 9 71 6 The system 1s equlvalent to Ax b whereA 7 5 and b 11 and the solut1on 1s x A b To compute this by hand the arithmetic is simpli ed by keeping the fraction ldetA in front of the matrix forA l The Study Guide cements on this in its discussion of Exercise 7 From Exercise 3 1 1 5 5 9 1 10 2 XA b Thusx12andx275 5 7 8 11 5 25 5 1 2 1 1 12 2 112 2 6 1 7 a or 5 12 1 12 2A5 5 1 2 5 1 25 5 71 1 12 2 1 1 l8 9 x A b1 S1m11ar calculatlons glve 2 5 1 3 2 8 4 A lb 11 A lb 6 A lb 13 2 5 3 2 4 5 Abbbbr12 1123 3912345123 565 12 112312 1123 0 2 8 10 4 10 014 5 2 5 0 l 9 11 6 13 Uquot 4 5 2 5 9 11 6 13 The solutlons are 4 5 2 and 5 the same as 1n part a 90 CHAPTER 2 M atrix Algebra Note The Study Guide also discusses the number of arithmetic calculations for this Exercise 7 stating that 1 whenA is large the method used in b is much faster than usingA p A DJ 8 Leftmultiply each side of the equation AD 1 by A 1 to obtain H N A Equot A IAD A II ID A l and D A l Parentheses are routinely suppressed because of the associative property of matrix multiplication a True by de nition of invertible b False See Theorem 6b 1 l c False If A 0 0 then ab 7 ed l 7 0 0 but Theorem 4 shows that this matrix is not invertible because ad7 be 0 d True This follows from Theorem 5 which also says that the solution ofo b is unique for each h e True by the box just before Example 6 21 False The product matrix is invertible but the product of inverses should be in the reverse order See Theorem 6b b True by Theorem 6a c True by Theorem 4 d True by Theorem 7 e False The last part of Theorem 7 is misstated here The proof can be modeled after the proof of Theorem 5 The nX p matrix B is given but is arbitrary SinceA is invertible the matrix A lB satisfies AX B becauseAA 1B A A lB IB B To show this solution is unique letXbe any solution ofAX B Then leftmultiplication of each side by A 1 shows thatXmust be A lB A 1 AX A IB IXA 1B and XA 1B If you assign this exercise consider giving the following Hint Use elementary matrices and imitate the proof of Theorem 7 The solution in the Instructor s Edition follows this hint Here is another solution based on the idea at the end of Section 22 Write B b1 bp andX u1 up By definition of matrix multiplication AX Aul Aup Thus the equationAX B is equivalent to thep systems Aul b1 Aup bp SinceA is the coefficient matrix in each system these systems may be solved simultaneously placing the augmented columns of these systems next toA to form A b1 hp A B SinceA is invertible the solutions ul up are uniquely determined and A b1 bp must row reduce to I 111 up I X By Exercise llXis the unique solutionA lB ofAX B Leftmultiply each side of the equation AB AC by A 1 to obtain A IAB A IAC 1B IC and B C This conclusion does not always follow whenA is singular Exercise 10 of Section 21 provides a counterexample Rightmultiply each side of the equation B 7 C D 0 by D 1 to obtain B7CDD 1 OD I B7CI 0 B7C 0 and B C The box following Theorem 6 suggests what the inverse of ABC should be namely C 1B 1A71 To verify that this is correct compute ABC C 1B 1A 1 ABCU 1B 1A 1 ABIB IA 1 ABB IA 1 AIA 1 AA 1 1 and CHISHA 1 ABC CHB IA IABC CHBIIBC CHBHBC CHIC 01C 1 22 Solutions 91 16 Let C AB Then CBr1 ABBH so CB 1 A1 A This shows thatA is the product of invertible matrices and hence is invertible by Theorem 6 Note The Study Guide warns against using the formula AB 1 B IA 1 here because this formula can be used only when bothA and B are already known to be invertible 17 Rightmultiply each side of AB BC by B71 ABB 1 BCB I AI BCB I A BCB I 18 Leftmultiply each side of A PBP 1 by P4 P IA P IPBFI PIA IBFI BIA BB1 Then rightmultiply each side of the result by P P IAP BF 1P P IAP B F 1AP B 19 Unlike Exercise 17 this exercise asks two things Does a solution exist and what is it First nd what the solution must be if it exists That is supposeX satis es the equation C 1A X15r1 1 Leftmultiply each side by C and then rightmultiply each side by B C0194 XB 1 CI 1A 0 C A Xng CB A 0 CB Expand the left side and then subtractA from both sides AIXICB AXCB XCB7A If a solution exists it must be CB 7A To show that CB 7A really is a solution substitute it forX C A CB 7AB1 CHCB15r1 CHCBB 1 III Note The Study Guide suggests that students ask their instructor about how many details to include in their proofs After some practice with algebra an expression such as CO 1A XB 1 could be simpli ed directly to A X15H without rst replacing CC 1 by 1 However you may wish this detail to be included in the homework for this section 20 a Leftmultiply both sides of A iAX 1 X 1B be to see thatB is invertible because it is the product of invertible matrices b Invert both sides of the original equation and use Theorem 6 about the inverse of a product which applies becauseX 1 and B are invertible A iAXf1B 1 B IQFI 1 BIX ThenA AX B IX A B IX The product A B1Xis invertible becauseA is invertible Since X is known to be invertible so is the other factor A B l by Exercise 16 or by an argument similar to part a Finally A B l 1A A B l 1A B1XX Note This exercise is dif cult The algebra is not trivial and at this point in the course most students will not recognize the need to verify that a matrix is invertible 21 SupposeA is invertible By Theorem 5 the equationAx 0 has only one solution namely the zero solution This means that the columns of A are linearly independent by a remark in Section 17 22 SupposeA is invertible By Theorem 5 the equationAx b has a solution in fact a unique solution for each b By Theorem 4 in Section 14 the columns ofA span Rquot 23 SupposeA is n Xn and the equationAx 0 has only the trivial solution Then there are no free variables in this equation and soA has n pivot columns SinceA is square and the n pivot positions must be in different rows the pivots in an echelon form ofA must be on the main diagonal HenceA is row equivalent to the n Xn identity matrix o to N UI N CA 27 CHAPTER2 MatrixAlgebra If the equationAx b has a solution for each h in Rquot thenA has a pivot position in each row b Theorem 4 in Section 14 SinceA is square the pivots must be on the diagonal ofA It follows thatA is row equivalent to I By Theorem 7 A is invertible a b 0 0 x1 0 Suppose A and ad7 bc 0 Ifa b 0 then examine This has the c d c d x2 0 d solution x1 This solution is nonzero except when a b c d In that case howeverA is the c b zero matrix andAx 0 for every vector x Finally if a and b are not both zero set x2 Then a b b b b 0 AxZ a a a because wb da 0 Thus x2 is a nontrivial solution ofo 0 c d a cb da 0 So in all cases the equationAx 0 has more than one solution This is impossible whenA is invertible by Theorem 5 soA is not invertible d b a b da bc 0 D1v1de both s1des by ad7 be to get CA I c a c d 0 cb ad a b d bad bc 0 c d c a 0 cbda39 Divide both sides by adi ha The right side is I The left side is AC because 1 a b d b a b 1 d b AC ad bc c d c a c d ad bc c a a InterchangeA and B in equation 1 after Example 6 in Section 21 row BA row B A Then replace B by the identity matrix row A row IA row1IAA b Using part a when rows 1 and 2 of A are interchanged write the result as row2A row2I A row2I row1A row1I AA row11 A EA row3 A row3 1 AA row3 I Here E is obtained by interchanging rows 1 and 2 of I The second equality in is a consequence of the fact that row EA row E A c Using part a when row 3 of A is multiplied by 5 write the result as row1 A row1I AA row1I row2A row2I A row2I A EA 5Arow3A 5Arow3I A 5 row3I Here E is obtained by multiplying row 3 of by 5 When row 3 of A is replaced by row3A 7 4 row1A write the result as row1A row2A row3A 4 row1A row1IAA row2I A row3I A 4 row1I AA 22 Solutions 93 row1IAA row11 row21 A row21 AEA row3I 4 row11 A row3I 4Arow11 Here E is obtained by replacing row3I by row3I N 4 row11 29AI12101210121010 72 3947010 l 41014 1014 l 30AI510101215012150 39 47010 1 451 l 2 15 0 l 0 75 2 A71 75 2 0 1 45 1 0 1 45 139 45 1 l 0 2 l 0 0 l 0 2 l 0 0 31 A I 3 l 4 0 l 0 N 0 l 2 3 l 0 2 N3 4 0 0 l 0 N3 8 N2 0 l l 0 2 l 0 0 l 0 0 8 3 l N 0 1 N2 3 l 0 N 0 l 0 10 4 l 0 0 2 7 3 l 0 0 2 7 3 l l 0 0 8 3 l 8 3 l N 0 l 0 10 4 l A71 10 4 l 0 0 1 72 32 l2 72 32 U2 2 l l 0 0 l 2 l l 0 0 32 A I 4 N7 3 0 l 0 N 0 l l 4 l 0 N2 6 N4 0 0 l 0 2 N2 2 0 l l 2 l l 0 0 N 0 l l 4 l 0 The matriXA is not invertible u u r Q to l l o and forj l n let a b and e denote thejth columns ofA B 0 0 l l and respectively Note that for j l n N l a am e because a and 2111 have the same entries except for the jth row by e N em and an n en To show thatAB I it suf ces to show thatAbj e for eachj Forj l n N 1 Ab Aej N em Aej NAej1 a N 2111 e 94 CHAPTER2 MatrixAlgebra andAbn Aen aH e Next observe that a e en for eachj Thus BajBejeKbjHbn 9 ej 91H 91H enrl en en 9 This proves that BA 1 Combined with the rst part this proves that B A71 Note Students who do this problem and then do the corresponding exercise in Section 24 will appreciate the Invertible Matrix Theorem partitioned matrix notation and the power of a proof by induction 34 Let 1 0 0 0 1 0 0 0 1 2 0 0 12 12 0 A 1 2 3 0 andB 0 13 13 1 2 3 n 0 0 ln 1n and for j l n let a b and e denote the jth columns of A B and respectively Note that for 11n71ajejAeblej 1 J J 1e l and b en To show thatAB 1 it suffices to show thatAbj e for eachj Forj 1 n71 1 1 1 AbjA7eJ me lj 7a maj1 ej en 1 en AlsoAb1 Alen la en Finally forj 1 n the sum b bquot is a telescoping sum n n whose value is 1e Thus J Ba JrBe a a Be 10 a a a b in e J which proves that BA 1 Combined with the first part this proves that B A l Note If you assign Exercise 34 you may wish to supply a hint using the notation from Exercise 33 Express each column of A in terms of the columns e1 en of the identity matrix Do the same for B 35 Row reduce A e3 2 7 9 0 1 3 4 1 1 3 4 1 1 3 4 1 602 5 600 1 2 2N0 1 2 2 4 1 2 7 9 0 0 1 1 2 0 0 1 4 1 3 0 15 1 3 0 15 1 0 0 3 0 1 400140014 Answer The third column of A 1 is 6 43 Ch 43 l DJ on There are many possibilities for C but C Write AD Ad1 d2 Adl Adz The structure ofA shows thatD 0 yDf 22 Solutions 95 M Write B A F where F consists of the last two columns of 13 and row reduce 32 92 4336 4392 25 9 546 180 154 50 27 0 0 1 0 0 537 1 0 N 0 1 0 149 0 1 0 0 1 683 69 15000 45000 The last two columns of A 1 are 72 1667 2195000 226667 690000 B 1 1 1 1 1 0 is the only one whose entries are 1 71 and 0 With only three possibilities for each entry the construction of C can be done by trial and error This is probably faster than setting up a system of 4 equations in 6 unknowns The fact thatA cannot be invertible follows from Exercise 25 in Section 21 becauseA is not square 1 0 0 0 0 0 1 There are 25 possibilities for D if entries of D are allowed to be 1 71 and 0 There is no 4gtlt2 matrix C such that CA I If this were true then CAx would equal x for all x in R4 This cannot happen because the columns of A are linearly dependent and soAx 0 for some nonzero vector x For such an x CAx C0 0 An alternate justification would be to cite Exercise 23 or 25 in Section 21 005 002 002 004 001 002 and 3 respectively 001 30 002 50 005 20 27 30 The de ections are 27 in 30 in and 23 in at points 1 2 23 M The sti ess matrix is D71 Use an inverse command to produce D 1 125 1 3 To find the forces in pounds required to produce a de ection of 04 cm at point 3 most students will use technology to solve Df 0 0 04 and obtain 0 75 10 Here is another method based on the idea suggested in Exercise 42 The first column of D 1 lists the forces required to produce a de ection of 1 in at point 1 with zero de ection at the other points Since the transformation y gt D ly is linear the forces required to produce a de ection of 04 cm at point 3 is given by 04 times the third column of D71 namely 04125 times 0 71 2 or 0 75 10 pounds To determine the forces that produce a de ections of 08 12 16 and 12 cm at the four points on the beam use technology to solve Df y where y 08 12 16 12 The forces at the four points are 12 15 215 and 12 newtons respectively 96 CHAPTER2 MatrixAlgebra 42 M To determine the forces that produce a de ection of 240 cm at the second point on the beam use technology to solve Df y where y 0 24 0 0 The forces at the four points are 7104 167 7113 and 560 newtons respectively to three signi cant digits These forces are 24 times the entries in the second column of D71 Reason The transformation y I gt Dily is linear so the forces required to produce a de ection of 24 cm at the second point are 24 times the forces required to produce a de ection of 1 cm at the second point These forces are listed in the second column of D71 Another possible discussion The solution of Dx 0 1 0 0 is the second column of D71 Multiply both sides of this equation by 24 to obtain D24x 0 24 0 0 So 24x is the solution of Df 0 24 0 0 The argument uses linearity but students may not mention this Note The Study Guide suggests using gauss swap bgauss and scale to reduce A 1 because I prefer to postpone the use of ref or rref until later If you wish to introduce ref now see the Study Guide s technology notes for Sections 28 or 43 Recall that Sections 28 and 29 are only covered when an instructor plans to skip Chapter 4 and get quickly to eigenvalues 23 SOLUTIONS Notes This section ties together most of the concepts studied thus far With strong encouragement from an instructor most students can use this opportunity to review and re ect upon what they have learned and form a solid foundation for future work Students who fail to do this now usually struggle throughout the rest of the course Section 23 can be used in at least three different ways 1 Stop after Example 1 and assign exercises only from among the Practice Problems and Exercises 1 to 28 I do this when teaching Course 3 described in the text s Notes to the Instructor If you did not cover Theorem 12 in Section 19 omit statements f and i from the Invertible Matrix Theorem 2 Include the subsection Invertible Linear Transformations in Section 23 if you covered Section 19 I do this when teaching Course 1 because our mathematics and computer science majors take this class Exercises 29740 support this material 3 Skip the linear transformation material here but discusses the condition number and the Numerical Notes Assign exercises from among 1728 and 41415 and perhaps add a computer project on the condition number See the projects on our web site I do this when teaching Course 2 for our engineers The abbreviation IMT here and in the Study Guide denotes the Invertible Matrix Theorem Theorem 8 5 7 1 The columns of the matrix 3 6 are not multiples so they are linearly independent By e in the IMT the matrix is invertible Also the matrix is invertible by Theorem 4 in Section 22 because the determinant is nonzero N 4 6 The fact that the columns of 6 9 are multiples is not so obvious The fastest check in this case may be the determinant which is easily seen to be zero By Theorem 4 in Section 22 the matrix is not invertible 3 Row reduction to echelon form is trivial because there is really no need for arithmetic calculations 5 0 0 5 0 0 5 0 0 3 7 0 N 0 7 0 N 0 7 0 The 3gtlt3 matrix has 3 pivot positions and hence is 8 5 1 0 5 1 0 0 1 invertible by c of the IMT Another explanation could be given using the transposed matrix But see the note below that follows the solution of Exercise 14 A 939 9 9 Thematrix 3 0 23 Solutions 97 7 0 4 l obviously has linearly dependent columns because one column is zero and 2 0 9 so the matrix is not invertible or singular by e in the IMT 7 0 3 5 1 0 2 1 0 2 1 0 2 1 0 2 N 0 3 5 N 0 3 5 N 0 3 5 7 4 9 7 4 9 7 0 9 15 0 0 0 The matrix is not invertible because it is not row equivalent to the identity matrix 1 5 4 1 5 4 1 5 4 0 3 40 3 40 3 4 7 3 6 0 0 9 12 0 0 0 The matrix is not invertible because it is not row equivalent to the identity matrix 1 301 1 301 1 301 358 3 0 480 0 480 2 632N0030N0030 0 1210 1210001 The 4gtlt4 matrix has four pivot positions and so is invertible by c of the IMT 1 3 7 4 0 5 9 6 The 4gtlt4 matr1x 0 0 2 8 1s 1nve1t1ble because 1t has four p1vot pos1t10ns by c of the IMT 0 0 0 10 4 0 7 7 1 2 3 1 1 2 3 1 M 6 1 11 9 6 1 11 9 0 11 7 15 7 5 10 19 7 5 10 19 0 9 31 12 1 2 3 1 4 0 7 7 0 8 5 11 1 2 3 1 1 2 3 1 1 2 3 1 0 8 5 11 0 8 5 11 0 8 5 11 0 9 31 12 0 0 25375 24375 0 25375 24375 0 ll 7 15 0 0 1250 1250 0 0 l 1 1 2 3 1 l 2 3 l 0 8 5 11 0 8 5 ll 0 0 1 1 0 0 1 1 0 0 25375 24375 0 0 0 1 The 4gtlt4 matrix is invertible because it has four pivot positions by c of the IMT 98 CHAPTER2 MatrixAlgebra 5 3 1 7 9 5 3 1 7 9 6 4 2 8 8 0 4 8 4 188 10 M 7 5 3 10 9 N 0 8 l6 2 36 9 6 4 9 5 0 6 22 216 212 8 5 2 11 4 0 2 4 2 104 5 3 1 7 9 5 3 1 7 9 0 4 8 4 188 0 4 8 4 188 N 0 0 0 1 34 N 0 0 1 21 7 0 0 l 21 7 0 0 0 1 34 0 0 0 0 l 0 0 0 0 l The 5gtlt5 matrix is invertible because it has ve pivot positions by c of the IMT H H a True by the IMT If statement d of the IMT is true then so is statement b b True If statement h of the IMT is true then so is statement e 0 False Statement g of the IMT is true only for invertible matrices d True by the IMT If the equationAx 0 has a nontrivial solution then statement d of the IMT is false In this case all the lettered statements in the IMT are false including statement c which means thatA must have fewer than n pivot positions e True by the IMT IfAT is not invertible then statement 1 of the IMT is false and hence statement a must also be false 21 True If statement k of the IMT is true then so is statement j b True If statement e of the IMT is true then so is statement h 0 True See the remark immediately following the proof of the IMT d False The rst part of the statement is not part i of the IMT In fact if A is any ngtltn matrix the linear transformation x I gtAx maps Rquot into Rquot yet not every such matrix has n pivot positions e True by the IMT If there is a b in Rquot such that the equationAx b is inconsistent then statement g of the IMT is false and hence statement f is also false That is the transformation x I gtAx cannot be onetoone N Note The solutions below for Exercises 1330 refer mostly to the IMT In many cases however part or all of an acceptable solution could also be based on various results that were used to establish the IMT 13 If a square upper triangular n Xn matrix has nonzero diagonal entries then because it is already in echelon form the matrix is row equivalent to In and hence is invertible by the IMT Conversely if the matrix is invertible it has n pivots on the diagonal and hence the diagonal entries are nonzero 14 If A is lower triangular with nonzero entries on the diagonal then these n diagonal entries can be used as pivots to produce zeros below the diagonal Thus A has n pivots and so is invertible by the IMT If one of the diagonal entries inA is zeroA will have fewer than n pivots and hence be singular Notes For Exercise 14 another correct analysis of the case whenA has nonzero diagonal entries is to apply the IMT or Exercise 13 toAT Then use Theorem 6 in Section 22 to conclude that since A7 is invertible so is its transpose A You might mention this idea in class but I recommend that you not spend much time discussingAT and problems related to it in order to keep from making this section too lengthy The transpose is treated infrequently in the text until Chapter 6 If you do plan to ask a test question that involves AT and the IMT then you should give the students some extra homework that develops skill using AT For instance in Exercise 14 replace columns by rows 23 Solutions 99 Also you could ask students to explain why an an matrix with linearly independent columns must also have linearly independent rows 15 9 9 gt0 N p n N N N 43 N UI N CA N I If A has two identical columns then its columns are linearly dependent Part e of the IMT shows that A cannot be invertible Part h of the IMT shows that a 5gtlt5 matrix cannot be invertible when its columns do not span R5 If A is invertible so is A71 by Theorem 6 in Section 22 By e of the IMT applied to A71 the columns of A 1 are linearly independent By g of the IMT C is invertible Hence each equation Cx v has a unique solution by Theorem 5 in Section 22 This fact was pointed out in the paragraph following the proof of the IMT By e of the IMT D is invertible Thus the equation Dx b has a solution for each h in R7 by g of the IMT Even better the equation Dx b has a unique solution for each h in R7 by Theorem 5 in Section 22 See the paragraph following the proof of the IMT By the box following the IMT E and F are invertible and are inverses So FE I EF and so E and F commute The matrix G cannot be invertible by Theorem 5 in Section 22 or by the box following the IMT So h of the IMT is false and the columns of G do not span Rquot Statement g of the IMT is false for H so statement d is false too That is the equation Hx 0 has a nontrivial solution Statement b of the IMT is false for K so statements e and h are also false That is the columns of K are linearly dependent and the columns do not span Rquot No conclusion about the columns of L may be drawn because no information about L has been given The equation Lx 0 always has the trivial solution Suppose thatA is square and AB I Ther1A is invertible by the k of the IMT Leftmultiplying each side of the equation AB I by A4 one has A IAB A II 1B A l and B A l By Theorem 6 in Section 22 the matrixB which is A4 is invertible and its inverse is A 1 1 which is A If the columns of A are linearly independent then sinceA is squareA is invertible by the IMT SoAZ which is the product of invertible matrices is invertible By the IMT the columns of A2 span Rquot Let Wbe the inverse ofAB Then ABW I andABW I SinceA is square A is invertible by k of the IMT Note The Study Guide for Exercise 27 emphasizes here that the equationAB W I by itself does not show thatA is invertible Students are referred to Exercise 38 in Section 22 for a counterexample Although there is an overall assumption that matrices in this section are square I insist that my students mention this fact when using the IMT Even so at the end of the course I still sometimes nd a student who thinks that an equation AB 1 implies thatA is invertible 28 Let Wbe the inverse ofAB Then WAB I and WAB I By 139 of the IMT applied to B in place ofA the matrix B is invertible 100 N 0 DJ p A CHAPTER 2 Matrix Algebra Since the transformation x I gtAx is not onetoone statement f of the IMT is false Then i is also false and the transformation x I gtAx does not map Rquot onto Rquot AlsoA is not invertible which implies that the transformation x I gtAx is not invertible by Theorem 9 Since the transformation x I gtAx is onetoone statement f of the IMT is true Then i is also true and the transformation x I gtAx maps Rquot onto Rquot Also A is invertible which implies that the transformation x I gtAx is invertible by Theorem 9 Since the equationAx b has a solution for each h the matrixA has a pivot in each row Theorem 4 in Section 14 SinceA is square A has a pivot in each column and so there are no free variables in the equationAx b which shows that the solution is unique Note The preceding argument shows that the square shape of A plays a crucial role A less revealing proof is to use the pivot in each row and the IMT to conclude thatA is invertible Then Theorem 5 in Section 22 shows that the solution of Ax b is unique 3 N 43 43 243 A If Ax 0 has only the trivial solution thenA must have a pivot in each of its n columns SinceA is square and this is the key point there must be a pivot in each raw ofA By Theorem 4 in Section 14 the equationAx b has a solution for each h in Rquot Another argument Statement d of the IMT is true soA is invertible By Theorem 5 in Section 22 the equationAx b has a unique solution for each h in Rquot 5 9 Solution in Study Guide The standard matrix of T is A 4 7 which is invertible because detA 0 By Theorem 9 the transformation T is invertible and the standard matrix of T1 is A71 From 7 9 the formula for a 2gtlt2 inverse A71 4 5 So 7 9 T 1x1x24 7x1 9x2 4x1 5x2 x2 6 8 The standard matrix of T is A 5 7 which is invertible because detA 2 0 By Theorem 9 7 8 T is invertible and T 1x Bx where B A71 5 6 Thus 1 7 8 x1 7 5 T1 4 3 x1 x2 2L 4L2 2x1 x2 2x1 x2 Solution in Study Guide To show that T is onetoone suppose that T u T v for some vectors u and v in Rquot Then STu STv where S is the inverse of T By Equation 1 u STu and STv v so u v Thus T is onetoone To show that T is onto suppose y represents an arbitrary vector in Rquot and define x Sy Then using Equation 2 T x T Sy y which shows that T maps Rquot onto Rquot Second proof By Theorem 9 the standard matrixA of T is invertible By the IMT the columns ofA are linearly independent and span R By Theorem 12 in Section 19 T is onetoone and maps Rquot onto Rquot If T maps R onto Rquot then the columns of its standard matrixA span Rquot by Theorem 12 in Section 19 By the IMT A is invertible Hence by Theorem 9 in Sectron 23 T is invertible and A 1 is the standar matrix of T1 Since A 1 is also invertible by the IMT its columns are linearly independent and span Rquot Applying Theorem 12 in Section 19 to the transformation T1 we conclude that T1 is a onetoone mapping of Rquot onto Rquot 43 l 40 4 H 23 Solutions 101 LetA andB be the standard matrices of T and U respectively ThenAB is the standard matrix of the mapping x I gt T U x because of the way matrix multiplication is de ned in Section 2 1 By hypothesis this mapping is the identity mapping soAB I SinceA and B are square they are invertible by the IMT and B A71 Thus BA I This means that the mapping x I gt U T x is the identity mapping ie UTx x for all x in Rquot LetA be the standard matrix of T By hypothesis T is not a onetoone mapping So by Theorem 12 in Section 19 the standard matrixA of T has linearly dependent columns SinceA is square the columns ofA do not span Rquot By Theorem 12 again T cannot map Rquot onto Rquot Given any v in Rquot we may write v T x for some x because T is an onto mapping Then the assumed properties of S and U show that Sv STx x and Uv UTx x So Sv and Uv are equal for each v That is S and U are the same function from Rquot into Rquot Given u v in Rquot let X Su and y SV Then TXTSu u and Ty TSV v by equation 2 Hence Su v STX Ty STx y BecauseT is linear x y By equation 1 Su Sv S0 S preserves sums For any scalar r Sru SrTx STrx BecauseTis linear rx By equationl rSu So S preserves scalar multiples Thus S ia a linear transformation M a e exact solution of 3 is x1 394 and x2 49 The exact solution of 4 is x1 290 and x2 7 200 b When the solution of 4 is used as an approximation for the solution in 3 the error in using the value of 290 for x1 is about 26 and the error in using 20 for x2 is about 308 O The condition number of the coef cient matrix is 3363 The percentage change in the solution from 3 to 4 is about 7700 times the percentage change in the right side of the equation This is the same order of magnitude as the condition number The condition number gives a rough measure of how sensitive the solution of Ax b can be to changes in b Further information about the condition number is given at the end of Chapter 6 and in Chapter 7 Note See the Study Guide s MATLAB box or a technology appendix for information on condition number Only the TI83 and TI89 lack a command for this 42 M MATLAB gives condA 23683 which is approximately 104 If you make several trials with MATLAB which records 16 digits accurately you should nd that x and x1 agree to at least 12 or 13 signi cant digits So about 4 signi cant digits are lost Here is the result of one experiment The vectors were all computed to the maximum 16 decimal places but are here displayed with only four decimal places 9501 38493 9501 21311 55795 2311 x rand4l b Ax The MATLAB solution 1s x1 Ab 6068 207973 6068 4860 8467 4860 102 CHAPTER2 MatrixAlgebra 0171 4858 712 However x 7 x1 2360 gtlt10 The computed solutlon X1 1s accurate to about 2456 12 decimal places M MATLAB gives condA 68622 Since this has magnitude between 104 and 105 the estimated accuracy of a solution of Ax b should be to about four or ve decimal places less than the 16 decimal places that MATLAB usually computes accurately That is one should expect the solution to be accurate to only about 11 or 12 decimal places Here is the result of one experiment The vectors were all computed to the maximum 16 decimal places but are here displayed with only four decimal places 4 DJ 2190 150821 2190 0470 8165 0470 x rand5l 6789 b Ax 190097 The MATLAB solution is x1 Ab 6789 6793 58188 6793 9347 145557 9347 3165 6743 However x x1 3343 X10711 The computed solution x1 is accurate to about 11 decimal places 0158 0005 44 M SolveAx 0 0 0 0 l MATLAB shows that condA 48gtlt105 Since MATLAB computes numbers accurately to 16 decimal places the entries in the computed value of x should be accurate to at least 11 digits The exact solution is 630 712600 56700 788200 44100 45 M Some versions of MATLAB issue a warning when asked to invert a Hilbert matrix of order 12 or larger using oatingpoint arithmetic The product1A71 should have several offdiagonal entries that are far from being zero If not try a larger matrix Note All matrix programs supported by the Study Guide have data for Exercise 45 but only MATLAB and lVIaple have a single command to create a Hilbert matrix The HP48G data for Exercise 45 contain a program that can be edited to create other Hilbert matrices Notes The Study Guide for Section 23 organizes the statements of the Invertible lVIatrix Theorem in a table that imbeds these ideas in a broader discussion of rectangular matrices The statements are arranged in three columns statements that are logically equivalent for any an matrix and are related to existence concepts those that are equivalent only for any an matrix and those that are equivalent for any nxp matrix and are related to uniqueness concepts Four statements are included that are not in the text s of cial list of statements to give more symmetry to the three columns You may or may not wish to comment on them I believe that students cannot fully understand the concepts in the IMT if they do not know the correct wording of each statement Of course this knowledge is not sufficient for understanding The Study Guide s Section 23 has an example of the type of question I often put on an exam at this point in the course The section concludes with a discussion of reviewing and re ecting as important steps to a mastery of linear algebra 24 Solutions 103 24 SOLUTIONS Notes Partitioned matrices arise in theoretical discussions in essentially every eld that makes use of matrices The Study Guide mentions some examples with references Every student should be exposed to some of the ideas in this section If time is short you might omit Example 4 and Theorem 10 and replace Example 5 by a problem similar to one in Exercises 1710 A sample replacement is given at the end of these solutions Then select homework from Exercises 1713 15 and 217 24 The exercises just mentioned provide a good environment for practicing matrix manipulation Also students will be reminded that an equation of the form AB I does not by itself makeA or B invertible The matrices must be square and the IMT is required 1 Apply the rowcolumn rule as if the matrix entries were numbers but for each product always write the entry of the left blockmatrix on the left I 0 A B IA0C IB0D A B E 1 C D EAIC EBID EAC EBD Apply the rowcolumn rule as if the matrix entries were numbers but for each product always write the entry of the left blockmatrix on the left E 0 A BEA0C EB0DEA EB 0 F C D 0AFC OBFD FC FD Apply the rowcolumn rule as if the matrix entries were numbers but for each product always write the entry of the left blockmatrix on the left 0WX0WIY 0XZY Z 1 0 Y Z IW0Y IXOZW X 4 Apply the rowcolumn rule as if the matrix entries were numbers but for each product always write the entry of the left blockmatrix on the left I 0 A B IA0C IB0D A B X 1 C D QlIC XBID QlC XBD 5 Compute the left side of the equation A B 1 0 AIBX A0BY C 0 X Y CI0X C00Y Set this equal to the right side of the equation ABX BY 0 I ABX0 BYI so that N S43 C 0 C Z 0 0 Since the 2 1 blocks are equal Z C Since the l 2 blocks are equal BY I To proceed further assume that B and Y are square Then the equation BY 1 implies that B is invertible by the IMT and Y B1 See the boxed remark that follows the IMT Finally from the equality of the l 1 blocks BX7A BIBXBI7A and X7BHA The order of the factors forX is crucial Note For simplicity statements 139 and k in the Invertible Matrix Theorem involve square matrices C and D Actually if A is HM and if C is any matrix such thatAC is the an identity matrix then C must be an too For AC to be defined C must have n rows and the equationAC 1 implies that C has n columns Similarly DA 1 implies thatD is ngtltn Rather than discuss this in class I expect that in Exercises 578 when 104 6 co CHAPTER2 MatrixAlgebra students see an equation such as BY I they will decide that both B and Y should be square in order to use the I MT Compute the left side of the equation X 0 A 0Ql0B X00C Ql 0 Y Z B C YAZB Y0ZC YAZB ZC Set this equal to the right side of the equation Ql 0 I 0 Ql I 0 0 so that YAZB ZC 0 I YAZB0 ZCI To use the equality of the l 1 blocks assume thatA andX are square By the IMT the equation Ql 1 implies thatA is invertible andX A l See the boxed remark that follows the IMT Similarly if C and Z are assumed to be square then the equation ZC I implies that C is invertible by the IMT and Z C 1 Finally use the 2 1 blocks and rightmultiplication by A71 YA iZB le YAA 1 leA 1 and Y HBA 1 The order of the factors for Y is crucial Compute the left side of the equation A Z X 0 0 0 Ql00B XZ0OI Y 0 I YA0IB YZ0II Set this equal to the right side of the equation Ql XZ I 0 Ql I XZ 0 so that YAB YZI 0 I YAB0 YZII To use the equality of the l 1 blocks assume thatA andX are square By the IMT the equationQl 1 implies thatA is invertible andX A71 See the boxed remark that follows the IMT Also X is invertible SinceXZ 0X 1XZ XIO 0 so Zmust be 0 Finally from the equality ofthe 2 1 blocks YA 7B Rightmultiplication byA 1 shows that YAA 1 iBA 1 and Y iBA l The order of the factors for Y is crucial Compute the left side of the equation A B X Y ZAXB0 AYB0 AZBI 0 I 0 0 I 0XIO 0YIO OZII Set this equal to the right side of the equation AX AY AZ B I 0 0 0 0 1 0 0 1 To use the equality of the l 1 blocks assume thatA andX are square By the IMT the equationQl 1 implies thatA is invertible andX A l See the boxed remark that follows the IMT SinceAY 0 from the equality of the l 2 blocks leftmultiplication by A 1 gives A IAY A710 0 so Y 0 Finally from the l 3 blocks AZ 7B Leftmultiplication byA 1 gives A IAZ A 17B and Z iAilB The order of the factors for Z is crucial Note The Study Guide tells students Problems such as 5710 make good exam questions Remember to mention the IMT when appropriate and remember that matrix multiplication is generally not commutative When a problem statement includes a condition that a matrix is square I expect my students to mention this fact when they apply the IMT 24 Solutions 105 9 Compute the left side of the equation I 0 0 A11 A12 IA110A21 0A31 IA12 0A22 0A32 X I 0 A21 A22 29111 IA21 0A31 29112 IA22 0A32 Y 0 1 A31 A32 YA11 0A21IA31 YA12 0A22 IA32 Set this equal to the right side of the equation A11 A12 Bu BIZ M11 1421 M12 1422 0 En YA11A31 YA12 A32 0 B32 A11 B11 A12 B12 so that X411A21 0 Ql12 A22 B22 YA11A310 YA12 A32 B32 Since the 21 blocks are equal Ql11 A21 0 and Ql11 A21 SinceAn is invertible right multiplication by A131 givesX A21A 1 Likewise since the 31 blocks are equal YA11 A31 0 and YA11 A31 Since A11 is invertible right multiplication by A131 gives Y A31A 1 Finally from the 22 entries XA12 A22 B22 Since X A21A1 11B22 A21A1 11A12 A22 10 Since the two matrices are inverses 7 0 0 T 0 0 I 0 0 C I 0 Z I 0 0 I 0 7A B I X Y I 0 0 1 Compute the left side of the equation 71 0 0 I 0 0 H0Z0X 10010Y 100001 C I 0 Z I 0 CIIZ0X C0H0Y C01001 7A B I X Y I AIBZIX A0BIIY A0B0H Set this equal to the right side of the equation 7 I 0 0 I 0 0 CZ I 0 0 I 0 7ABZX BY I 0 0 I 1 00 00 sothat CZ0 1 00 ABZX0 BY0 1 Since the 21 blocks are equal C Z 0 and Z C Likewise since the 3 2 blocks are equal B Y 0 and Y B Finally from the 31 entries A BZ X 0 and X A BZ SinceZ C X A B C ABC 11 a True See the subsectionAddition and ScalarMultiplication b False See the paragraph before Example 3 12 a True See the paragraph before Example 4 b False See the paragraph before Example 3 106 CHAPTER2 Matrix Algebra 13 You are asked to establish an and only statement First supose thatA is invertible p A p A UI D E and let A 1 Then F G B 0 D EBD BEI 0 0 C F GCF CG0 1 Since B is square the equation BD I implies that B is invertible by the IMT Similarly CG I implies that C is invertible Also the equation BE 0 imples that E B710 0 Similarly F 0 Thus 1 B 0 1 D E B 1 0 A 0 C E G 0 0 1 This proves thatA is invertible only B and C are invertible For the if part of the statement suppose that B and C are invertible Then provides a likely candidate for A 1 which can be used to show that A is invertible Compute B0B 10BB 1 0IO 0 C 0 C l 0 004 0 1 Since A is square this calculation and the IMT imply thatA is invertible Don t forget this nal sentence Without it the argument is incomplete Instead of that sentence you could add the equation 3 li 2rle CECH El You are asked to establish an and only statement First suppose thatA is invertible Example 5 shows thatAu andAzz are invertible This proves thatA is invertible only zfAn All are invertible For the part of this statement suppose thatAn andAzz are invertible Then the formula in Example 5 provides a likely candidate for A71 which can be used to show thatA is invertible Compute A11 A12 A131 A11A12A AHA i11 A120 A11A 12 A A12 A 0 A22 0 Ag 0AA220 0 A1 11A12A A22A 1 A11AA12AA12A 0 I l Since A is square this calculation and the IMT imply thatA is invertible ArzAgi l39ArzAgi I 0 I 0 I Compute the right side of the equation I 01A11 OM Y A11 01 y A11 AMY X I 0 S 0 I XA11 S 0 I XA11 XA11YS Set this equal to the left side of the equation A12 14111411 A11YA12 A11 AIIY A11 sothat XA11 XA11YS A21 A22 XA11A21 Since the l 2 blocks are equal AHY A12 SinceAn is invertible left multiplication by A111 gives XA11YSA22 Y Ail A12 Likewise since the 21 blocks are equal XAH AZI SinceAu is invertible right 9 gt1 p A on 24 Solutions 107 multiplication by A1 gives that X AZIA l One can check that the matrix S as given in the exercise satisfies the equation X AMY S A22 with the calculated values of X and Y given above Suppose thatA andAn are invertible First note that I 0 I 0 I 0 X 1 X 1 0 1 I Y I Y I 0 0 1 0 1 0 1 I 0 Since the matrices and X I and I Y 0 I are square they are both invertible by the MT Equation 7 may be left multipled by 1 0 1 1 Y 1 and r1ght mult1pled by to find X I 0 I All 0 1 0 1A1 Y4 0 S X 1 0 1 A11 0 Thus by Theorem 6 the matr1x 0 S 1s 1nvert1ble as the product of mvertlble matr1ces Fmally Exercise 13 above may be used to show that S is invertible The columnrow expansions of Gk and GM are Gk Xng c011Xkrow1X colkXkrowkXkT and Gk1 X k1X 51 0011Xk1r0W1X1 quot39 001 Xk rowk X121 c011m Xk r0wkl X121 0011Xkr0W1X quot39 00115095 ow4X 001k1Xk1rowk1X Gk 001k1 Xk1r0We1 X1 since the rst k columns of XM are identical to the first k columns of X k Thus to update Gk to produce GM the number colk XM rowk X g should be added to Gk Since WX x0 T XT XTX XTx0 W W X x0 x3 x3 X x3 x0 By applying the formula for S from Exercise 15 S may be computed S xix0 ngXTX 1XTx0 x3 1m XXTX 1XTx0 T XOMXO 108 19 N p A N N CHAPTER2 MatrixAlgebra The matrix equation 8 in the text is equivalent to A sInxBu 0 and Cxuy Rewrite the first equation as A SI x Bu When A S1 is invertible x A SI 1 Bu A SI ilBu Substitute this formula for x into the second equation above C A sIn 1Bu u y sothat 1 u CA s1 TlBu y Thus y 1m CA SI ilBu If Ws 1m CA SI ilB then y Wsu The matrix Ws is the Schur complement of the matrix A S1 in the system matrix in equation 8 The matrix in question is A BC SI B l c A By applying the formula for S from Exercise 15 S may be computed S 1m CA BC sjm B 1m CA BC sm 1B A2101010 0010 39a39 3 13 1 3 3 0 1201 00 1 0 0 A2 0 I I3 bM2A 0A 0A20 39 I AI AAA 0 Let C be any nonzero 2X3 matrix Define A I Then 2 C 2 I3 0 M13 0 130 00 3 0 A C 12 C 12 CI3 12C 0 122 0 12 The product of two lgtltl lower triangular matrices is lower triangular Suppose that for n k the product of two ka lower triangular matrices is lower triangular and consider any k lgtlt k 1 matrices A1 and B1 Partition these matrices as AZaOT BZbOT 1 IWB v A whereA and B are ka matrices v and w are in Rk and a and b are scalars SinceA1 and B1 are lower triangular so areA and B Then ab AB a 0T b of ab0Tw a0T0TB 0T 1 1 v A w B vbAw v0TAB bvAw AB SinceA and B are 10 AB is lower triangular The form ofAlB1 shows that it too is lower triangular Thus the statement about lower triangular matrices is true for n k l if it is true for n k By the principle of induction the statement is true for all n 2 l 24 Solutions 109 Note Exercise 23 is good for mathematics and computer science students The solution of Exercise 23 in the Study Guide shows students how to use the principle of induction The Study Guide also has an appendix on The Principle of Induction at the end of Section 24 The text presents more applications of induction in Section 32 and in the Supplementary Exercises for Chapter 3 1 0 0 0 1 0 0 1 1 0 0 1 1 0 24 LetAnl 1 l 0 B 0 1 1 0 1 1 1 1 0 1 1 By direct computation AZBZ 2 Assume that for n k the matrix AkBk is 1k and write 1 0T 1 0T A and B v A W Bk wherevandwareianvTl l landwT7l 0 0Then 1 0T 1 0T 107w 07 003 1 0T AklBk1 Ik1 v Ar w Bk vAkw v0TAkBf 0 1k The 21entry is 0 because v equals the rst column ofAk andAkw is 71 times the first column of Ak By the principle of induction Aan In for all n 3 2 Since A and Bquot are square the MT shows that these matrices are invertible and B A 1 Note An induction proof can also be given using pa1titions with the form shown below The details are slightly more complicated A 0 Bk 0 Am VT 1 and BM WT 1 A 0 B 0 AkBk0wT Ak00 I 0 AHIBHI VT 1 WT 1 vTBkwT vT0l 0T 1 21k The 21entry is 0T because vT times a column of Bk equals the sum of the entries in the column and all of such sums are zero except the last which is 1 So VTBk is the negative of WT By the principle of induction Aan In for all n 3 2 SinceAn and B are square the MT shows that these matrices are inve1tible and B A 1 25 First visualize a pa1tition of A as a 2gtlt2 blockaiiagonal matrix as below and then visualize the 22block itself as a blockdiagonal matrix That is 110 CHAPTER 2 Matrix Algebra 3 4 Observe thatB is invertible and B1 2 5 3 5 By Exercise 13 the block diagonal matrixAzz is invertible and 5 0 0 0 3 4 0 25 35 5 2 Next observe thatAn is also invertible with inverse 3 I By Exercise 13A itself is invertible and its inverse is block diagonal 5 2 0 0 0 1 3 1 0 0 0 71 All A 0 0 5 0 0 0 0 0 0 3 4 0 0 0 25 35 M This exercise and the next which involve large matrices are more appropriate for MATLAB Maple and Mathernatica than for the graphic calculators a Display the submatrix ofA obtained from rows 15 to 20 and columns 5 to 10 MATLAB A1520 510 Maple submatrixA 15 20 510 Mathematica Take A 1520 510 b Insert a 5gtlt10 matrixB into rows 10 to 14 and columns 20 to 29 ofmatrixA MATLAB A 10 14 20 29 B The semicolon suppresses output display Maple copyinto B A 10 2 0 The colon suppresses output display Mathematica For i10 ilt14 i For j20 jlt29 j A ij B i9 j19 Colonsuppresses output P A 0 To create B 0 T with MATLAB build B out of four blocks A B A zeros3020 zeros 2030 A39 Another method rst enter B A and then enlarge B with the command B2150 3150 A39 This places A7 in the 2 2 block of the largerB and lls in the 1 2 and 2 1 blocks with zeros For Maple B matrix50500 1 1 copyinto transposeA B For Mathernatica B BlockMatrix A ZeroMatrix3020 ZeroMatrix2030 TransposeA copyinto A B 21 31 24 Solutions 111 27 a M ConstructA from four blocks say C11 C12 C21 and C22 for example with C11 a 30x30 matrix and C22 a 20x20 matrix MATLAB C11 A130 130 B130 130 C12 A130 3150 B130 3150 C21 A3150 130 B3150 130 C22 A3150 3150 B3150 3150 C C11 C12 C21 C22 The commands in Maple and Mathematica are analogous but with different syntax The rst commands are Maple C11 submatrixA 1 30 1 30 submatrixB 1 30 1 30 Mathematica c11 Take A 130 1 3o 1 Take 3 130 1 3o 1 The algebra needed comes from block matrix multiplication AB 2 A11 A12B11 B12 1411311 141sz1 1411312 141sz2 A21 A22 BZI B22 A21B11 AZZBZI AZIBIZ A22B22 Partition bothA and B for example with 30x30 1 1 blocks and 20gtlt20 2 2 blocks The four necessary submatrix computations use syntax analogous to that shown for a Uquot P A 0 b The algebra needed comes from the block matrix equation 11 X1 1 where x1 and b1 A21 A22 X2 b2 are in R30 and x2 and b are in R20 ThenA1 1x1 b1 which can be solved to produce x1 Once x1 is found rewrite the equation Allxl A22x2 b2 asAzzxz c where c b2 7A21x1 and solve Azzxz c or x Notes The following may be used in place of Example 5 Example 5 Use equation to find formulas forX Y and Z in terms ofA B and C Mention any assumptions you make in order to produce the formulas X 0 I 0 I 0 Ck Y Z A B C 1 Solution This matrix equation provides four equations that can be used to fde Y and Z X 0 1 0 0 Y ZA C Y 0 ZB 1 Note the order of the factors The rst equation says thatX 1 To solve the fourth equation ZB I assume thatB and Z are square In this case the equation ZB 1 implies that B and Z are invertible by the IMT Actually it suffices to assume either that B is square or that Z is square Then rightmultiply each side of ZB I to get ZBJBY1 IB 1 and Z HI Finally the third equation is Y ZA C So Y B 1A C and Y CiB lA The following counterexample shows that Z need not be square for the equation above to be true 112 CHAPTER2 MatrixAlgebra Note that Z is not determined byA B and C when B is not square For instance another Z that works in 3 5 0 this counterexample is Z l 2 0 25 SOLUTIONS Notes Modern algorithms in numerical linear algebra are often described using matrix factorizations For practical work this section is more impoitant than Sections 47 and 54 even though matrix factorizations are explained nicely in terms of change of bases Computational exercises in this section emphasize the use of the LU factorization to solve linear systems The LU factorization is performed using the algorithm explained in the paragraphs before Example 2 and performed in Example 2 The text discusses how to build L when no interchanges are needed to reduce the given matrix to U An appendix in the Study Guide discusses how to build L in permuted unit lower triangular form when row interchanges are needed Other factorizations are introduced in Exercises 22726 1 0 0 3 7 2 7 1 L l l 0 U 0 2 l b 5 FirstsolveLyb 2 5 1 0 0 1 2 100 7100 7 L b l l 0 5 N 0 l 0 2 The only arithmetic is in column4 2 5 1 2 0 5 1 16 1 0 0 7 7 0 l 0 2 soy 2 0 0 1 6 6 Next solve Ux y using backsubstitution with matrix notation 3 7 2 7 3 7 2 7 3 7 0 19 U y 0 2 1 2 0 2 1 2 0 2 0 8 0 0 1 6 0 0 1 6 0 0 1 6 3 7 0 19 3 0 0 9 1 0 0 3 N 0 1 0 4 0 1 0 4 0 1 0 4 0 0 l 6 0 0 1 6 0 0 l 6 So x 3 4 To confirm this result row reduce the matrix A b 3 7 2 7 3 7 2 7 3 2 7 A b 3 5 1 5 0 2 1 2 0 1 2 6 4 0 2 0 10 4 16 0 0 l 6 7 2 From this point the row reduction follows that of U y above yielding the same result 25 Solutions 113 1 0 0 4 3 5 2 2 L 1 l 0 U 0 2 2 b 4 FirstsolveLyb 2 0 1 0 0 2 6 1 0 0 2 1 0 2 L b 1 1 0 4 0 1 2 2 0 1 6 0 0 2 2 soy 2 2 Next solve Ux y using backsubstitution with matrix notation 4 3 5 2 4 3 5 2 4 3 0 7 U y 2 2 2 0 2 2 2 0 2 0 4 0 2 2 0 0 1 1 0 0 1 1 3 0 7 4 0 0 1 1 0 0 14 1 0 2 0 1 0 2 0 1 0 2 0 1 1 0 0 1 1 0 0 1 1 so x l 42 1 To con rm this result row reduce the matrix A b 4 3 5 2 4 3 5 2 A b 4 5 7 4 0 2 2 2 8 6 8 6 0 0 2 2 From this point the row reduction follows that of U y above yielding the same result 1 0 0 2 1 2 1 3 L 3 l 0 U 0 3 4 b 0 FirstsolveLyb 4 1 1 0 0 1 4 1 0 0 1 1 0 0 1 1 0 0 1 Lb 310001030103 4 1 1 4 0 1 1 0 0 0 1 0 1 soy 3 3 Next solve Ux y using backsubstitution with matrix notation 2 1212 10 52 10 5 Uy0 3430 30 90 10 3 0 0 1 3 0 0 1 3 0 0 1 3 2 0 0 2 0 1 0 3 0 0 1 so x 71 3 3 114 CHAPTER2 MatrixAlgebra 1 0 0 2 2 4 0 4L12 1 0 5U 0 1b 5 FirstsolveLyb 32 5 1 0 0 6 7 1 0 0 0 1 0 0 0 1 0 0 0 Lb1210 5010 5010 5 32 5170 517001 18 so y 5 18 Next solve Ux y using backsubstitution with matrix notation 2 2 4 0 2 2 4 0 2 2 0 12 Uy0 2 1 50 2 1 50 20 2 0 0 6 18 0 0 1 3 0 0 1 3 2 2 0 12 2 0 0 10 1 0 0 5 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 3 0 0 1 3 0 0 1 sox75 13 1 0 0 0 1 2 4 3 1 2 1 0 0 0 3 1 0 7 5 L U h FirstsolveLyb 1 0 1 0 0 0 2 0 4 3 5 1 0 0 0 3 1 0 0 0 1 1 0 0 0 1 2 1 0 0 7 0 1 0 0 5 L b 1 0 1 0 0 0 0 1 0 1 4 3 5 1 3 0 3 5 1 7 1 0 0 0 1 1 0 0 0 1 0 1 0 0 5 0 1 0 0 5 0 0 1 0 1 0 0 1 0 1 0 0 5 1 8 0 0 0 1 3 1 5 so y 3 Next solve Ux y using backsubstitution with matrix notation 1 2 4 311 2 40 8 0 10 5 0 3 10 Uy 0 0 2110 0 20 4 0001 3 0001 3 Solutions 1 15 25 0001 0010 300 1000 i N i 2 4 0 8 3 1 0 5 0 1 0 2 0 0 1 3 1 N 0 0 0 72 71 2 73 200 1 0 0 1 0 0 0 SOX b First solve Ly 1 2 b 1 2 i 0201 4520 3300 L b 2 0 1 0 soy y using backsubstitution with matrix notation Next solve UK 350 00 20 2 00 20 2 U y 3 6 1 1 134011300 0350 10300 N0 01010 0 110001 3 72 1 1 300 310003 2 1 0 0 1 0 0 0 1 0 0 0 SOX 116 CHAPTER2 MatrixAlgebra 2 5 7 Place the first pivot column of 3 4 into L after dividing the column by 2 the pivot then add 32 times row 1 to row 2 yielding U Alel an Pilaf 2 72 i i 312 il in 312 8 Row reduceA to echelon form using only row replacement operations Then follow the algorithm in Example 2 to nd L 2H 9 i hf 6 l 1 10 L 231 231 1 2 3 1 2 3 1 2 9A 3 2100120 3 12U 0 9 560 20 0 2 1 73 73 78 i l i 1 1 0 0 1 1 L 1 l 0 25 Solutions 117 3 4 5 3 4 5 3 4 10 A 10 8 9 N 0 g 1 N 0 2 1U 15 1 2 0 10 14 0 0 9 E3 10 g le1 5 2 9 ill i l 1 0 0 2 1 L 2 10 3 5 1 3 5 1 11A 6 7 20 40 5 4U l3 1 l 1 12 Row reduceA to echelon form using only row replacement operations Then follow the algorithm in Example 2 to nd L Use the last column of 13 to make L unit lower triangular 422 422 42 A15 40 50 7 5U 7 6 2 4 0 14 10 0 0 0 4 2 iii ill 1 0 0 Matrix Algebra 118 CHAPTER2 U No more pivots 3 0 1 0 0 3 5 3 1 5 8 4 4 l 13 15 10 l i D 1 10 i Use the last two columns of I 4 to make L unit lower triangular 4 2 2 72 1 14 A 2 i Use the last two columns of I 4 to make L unit lower triangular 5 10 D 3 2 1 20 2 Solutions 1 19 25 U 4 4 2 2 44 2 0 5 303 53 610 1 00 0 2 0 44 2 6 97 3 1 48 0 15 A 16 A Use the last three columns of 15 to make L unit lower triangular 21 0 0000 1 1 2 32 1 3001 4 277 ll 1 21 2 21 L l 0 100 20 300 4 3 reduce L I 1 To nd L71 use the method of Section 22 that is row 2 17 L 0100100 100101101L 1 2 0 1 1 L 1 1 0 0 0 120 CHAPTER2 MatrixAlgebra 1 0 0 so L71 1 1 0 Likewiseto nd U71 rowreduce U1 2 0 1 4 3 5 1 0 0 4 3 0 1 0 52 UI0 2 20100 2001 1 0 0 2 0 0 1 0 0 2 0 0 1 4 0 0 1 32 1 1 0 0 14 38 14 0 2 0 0 1 1 0 1 0 0 12 12 1 02H 0 0 2 0 0 1 0 0 1 0 0 12 14 38 14 so U21 0 12 12 Thus 0 0 12 14 38 14 1 0 0 18 38 14 A 11J4171 0 12 12 1 1 0 32 12 12 0 0 12 2 0 1 1 0 12 1 0 0 2 1 2 18 L 3 1 0U 3 4 TofmdL 1rowreduceL I 4 1 1 0 0 1 1 0 0 1 0 0 1 0 0 1 0 0 L ns 1 0 0 1 0 No 1 0 3 1 0 4 1 1 0 0 1 0 1 1 4 0 1 0 1 0 0 N 0 3 1 0 E1 1 1 1 1 0 1 13 23 0 0 13 43 1 0 0 1 2 No 0 0 1 0 0 1 0 1 0 2 2 0 0 0 U3 46 No 1 1 0 0 1 0 0 1 0 0 U2 U6 U3 N 0 1 0 0 U3 4B U U4L 0 0 1 0 0 1 p A D N G 25 Solutions 121 12 16 13 so U 1 0 13 43Thus 0 0 1 12 16 13 1 0 0 13 12 13 A 1U 1L 1 0 13 43 3 1 0 73 1 43 0 0 1 1 1 1 1 1 1 LetA be a lowertriangular n X 71 matrix with nonzero entries on the diagonal and consider the augmented matrix A I a The 1 lentry can be scaled to l and the entries below it can be changed to 0 by adding multiples of row 1 to the rows below This affects only the rst column of A and the rst column of So the 2 2entry in the new matrix is still nonzero and now is the only nonzero entry of row 2 in the rst n columns becauseA was lower triangular The 2 2entry can be scaled to 1 the entries below it can be changed to 0 by adding multiples of row 2 to the rows below This affects only columns 2 and n 2 of the augmented matrix Now the 3 3 entry inA is the only nonzero entry of the third row in the rst n columns so it can be scaled to l and then used as a pivot to zero out entries below it Continuing in this wayA is eventually reduced to I by scaling each row with a pivot and then using only row operations that add multiples of the pivot row to rows below 9 The row operations just described only add rows to rows below so the I on the right in A 1 changes into a lower triangular matrix By Theorem 7 in Section 22 that matrix is A71 LetA LU be an LU factorization forA Since L is unit lower triangular it is invertible by Exercise 19 Thus by the Invertible Matrix Theroem L may be row reduced to I But L is unit lower triangular so it can be row reduced to I by adding suitable multiples of a row to the rows below it beginning with the top row Note that all of the described row operations done to L are rowreplacement operations If elementary matrices E1 E2 Ep implement these rowreplacement operations then EPE2E1A EP E2E1LU IU U This shows thatA may be row reduced to U using only rowreplacement operations Solution in Study Guide Suppose A BC with B invertible Then there exist elementary matrices E1 Ep corresponding to row operations that reduce B to I in the sense that Ep EIB I Applying the same sequence of row operations to A amounts to leftmultiplyingA by the product Ep E1 By associativity of matrix multiplication EPE1AEPEIBCICC so the same sequence of row operations reduces A to C First nd an LU factorization forA Row reduceA to echelon form using only row replacement operations 23 4 23 2 4 23 2 4 23 6 9 5 8 0 3 1 1 0 3 1 1 A2 7 3 90 3 1 60 0 0 4 2 2 1062 7000 5 6 3 3 4 0 9 3 13 0 0 010 122 CHAPTER2 MatrixAlgebra 2 4 2 3 0 3 l l N 0 0 0 0 0 0 then follow the algorithm in Example 2 to nd L Use the last two columns of 15 to make L unit lower triangular 4 6 5 7 6 9 10 T2 3 5 l l 7 l l 0 0 0 0 3 l 3 l 0 0 0 l l l L l l l 0 0 2 2 l l 2 2 l l 0 3 3 2 0 l 3 3 2 0 1 Now notice that the bottom two rows of U contain only zeros If one uses the rowcolumn method to nd LU the entries in the nal two columns of L will not be used since these entries will be multiplied zeros from the bottom two rows of U So let B be the rst three columns of L and let C be the top three rows of U That is l 0 3 l 0 2 4 2 3 B l l 1 C 0 3 l l 2 2 l 0 0 0 5 3 3 2 Then B and C have the desired sizes and BC LU A We can generalize this process to the case where A in m X n A LU and U has only three nonzero rows let B be the first three columns of L and let C be the top three rows of U 23 21 Express each row of D as the transpose of a column vector Then use the multiplication rule for partitioned matrices to write di ACDc1 c2 c3 c4 cldf czd2T c3d3T c4df di which is the sum of four outer products b SinceA has 400 X 100 40000 entries Chas 400 X 4 1600 entries and D has 4 X 100 400 entries to store C and D together requires only 2000 entries which is 5 of the amount of entries needed to storeA directly 24 N CA 25 Solutions 123 Since Q is square and QTQ I Q is invertible by the Invertible Matrix Theorem and Q 1 QT Thus A is the product of invertible matrices and hence is invertible Thus by Theorem 5 the equationAx b has a unique solution for all h FromAx b we have QRX b QTQRX QTb Rx QTb and nally x R IQTb A good algorithm for nding x is to compute QTb and then row reduce the matrix R QTb See Exercise 11 in Section 22 for details on why this process works The reduction is fast in this case because R is a triangular matrix A UDVT Since Uand V7 are square the equations UT UI and VT VI imply that Uand V7 are invertible by the IMT and hence U 1 U T and VT 1 V Since the diagonal entries 03 o n in D are nonzero D is invertible with the inverse of D being the diagonal matrix with 0171 039 on the diagonal Thus A is a product of invertible matrices By Theorem 6A is invertible andA 1 UDV T 1 IT 1D 1 U1 VD IUT If A PDF 1 where P is an invertible 3 X 3 matrix and D is the diagonal matrix 1 0 0 D 0 12 0 0 0 13 then A2 PDP 1PDP 1 PDP 1PDP 1 PDIDP I PDZP I andsince 1 0 0 1 0 0 1 0 0 1 0 0 D2012 0012 00122 0014 0 0 0130 013 0 0132 0 019 1 0 0 A2P 0 14 0 P 1 70 0 19 Likewise A3 PD3P 71 so 71 0 0 1 0 0 A3P 0 123 0 P 1P 0 18 0 P 1 0 0 133 0 0 127 In generalAk PDkP l so 1 0 0 AkP 0 12k 0 P 1 70 0 13k 27 First consider using a series circuit with resistance R1 followed by a shunt circuit with resistance R for the network The transfer matrix for this network is R 1 0 1 121 1 1 1122 1 0 1 1R2 R1R2R2 124 CHAPTER 2 29 Matrix Algebra For an input of 12 volts and 6 amps to produce an output of 9 volts and 4 amps the transfer matrix must satisfy 1 R1 12 12 6R1 9 1122 121 R Rji 6 i 2 912 6R1 6R2R2 i 2 4i Equate the top entries and obtain R1 ohm Substitute this value in the bottom entry and solve to obtain R2 ohms The ladder network is Next consider using a shunt circuit with resistance R1 followed by a series circuit with resistance R2 for the network The transfer matrix for this network is 1 R2 1 0 R1 R2R1 R2 0 1 1121 1 1121 1 For an input of 12 volts and 6 amps to produce an output of 9 volts and 4 amps the transfer matrix must satisfy R1 R2R1 R2 12 12R112R2R1 6R2 9 1121 1 6 12R16 4 Equate the bottom entries and obtain R1 6 ohms Substitute this value in the top entry and solve to obtain R2 ohms The ladder network is The three shunt circuits have transfer matrices 1 0 1 0 1 0 and 1R1 1 1R2 1 lR3 1 respectively To nd the transfer matrix for the series of circuits multiply these matrices 1 0 1 0 d 1 o 1 0 1123 1 1R2 1 an 1121 1 1R11R21R3 1 Thus the resulting network is itself a shunt circuit with resistance lR1 lR2 1R3 1 0 a The rst c1rcu1t 1s a shunt c1rcu1t w1th res1stance R1 ohms so 1ts transfer matr1x 1s NR 1 1 le 1 The second c1rcu1t 1s a ser1es c1rcu1t w1th res1stance R2 ohms so 1ts transfer matr1x 1s 0 25 Solutions 125 1 0 The third circuit is a shunt circuit with resistance R3 ohms so its transfer matrix is NR 1 3 The transfer matrix of the network is the product of these matrices in righttoleft order 1 0 1 122 1 0 121122121 122 1123 1 0 1 1121 1 R1R2R3R3 122123123 To nd a ladder network with a structure like that in part a and with the given transfer matrix A we must nd resistances R1 R2 and R3 such that A43 12 R1R2R1 R2 14 3 R1R2R3R3 R2R3R3 From the 1 2 entries R2 12 ohms The 1 1 entries now give R1 12R1 43 which may be solved to obtaian 36 ohms Likewise the 2 2 entries give R3 12R3 3 which also may be solved to obtain R3 6 ohms Thus the matrixA may be factored as A 1 0 1 122 1 0 1R3 1 0 1 1121 1 1 0 1 12 1 0 16 1 0 1 136 1 The ladder network is P The transfer matrix of this network is the product of the individual transfer matrices in right to left order 3 1311ij 113 R2 R3R2 R3 R1R2 R3R2 1R2 R1 R2R2 By setting the matrixA from the previous exercise equal to this matrix one may nd that R2R3R2 R3 R1R2R3R2 43 12 1R2 Rum122 izi m 3 i Set the 2 1 entries equal and obtain R2 4 ohms Substitute this value for R2 equating the 2 2 entries and solving gives R1 8 ohms Likewise equating the 1 1 entries gives R3 43 ohms 126 CHAPTER2 MatrixAlgebra The ladder network is Note The Study Guide 3 MATLAB box for Section 25 suggests that for most LU factorizations in this section students can use the gauss command repeatedly to produce U and use paper and mental arithmetic to write down the columns of L as the row reduction to U proceeds This is because for Exercises 7716 the pivots are integers and other entries are simple fractions However for Exercises 31 and 32 this is not reasonable and students are expected to solve an elementary programming problem The Study Guide provides no hints 31 M Store the matrixA in a temporary matrix B and create L initially as the 8X8 identity matrix The following sequence of MATLAB commands lls in the entries of L below the diagonal one column at a time until the rst seven columns are filled The eighth column is the nal column of the identity matrix L28 1 B28 1 B1 1 B gauss B 1 L38 2 B38 2B2 2 B gauss B 2 L88 7 B88 7B7 7 U gauss B7 Of course some students may realize that a loop will speed up the process The for end syntax is illustrated in the MATLAB box for Section 56 Here is a MATLAB program that includes the initial setup ofB andL B A L eye 8 for j 1 7 Lj18 j Bj18 jBj j B gauss B j end U B a To four decimal places the results of the LU decomposition are 1 0 0 0 0 0 0 0 25 1 0 0 0 0 0 0 25 0667 1 0 0 0 0 0 0 2667 2857 1 0 0 0 0 L 0 0 2679 0833 l 0 0 0 0 0 0 29 17 2921 1 0 0 0 0 0 0 2697 0861 l 0 0 0 0 0 0 2948 2931 l b The result of solving Ly b and then Ux y is 1 0 25 1 37333 10667 0 34286 0 0 0 0 0 0 0 0 1 2857 37083 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 10833 1 0 33919 2921 1 0 37052 10861 0 0 33868 x5956965885423921397156029876089411512043n 2953 0866 0945 0509 0318 0227 0010 0082 c 2471 0866 2953 0509 0945 0227 0318 0082 0100 cgt 0 mm 0945 0509 3271 1093 1045 0591 0318 0227 m o N p cgt cgt c O ale m 55 0509 0945 1093 3271 0591 1045 0227 0318 b IOOOO 0318 0227 1045 0591 3271 1093 0945 0509 0227 0318 0591 1045 1093 3271 0509 0945 0010 0082 0318 0227 0945 0509 2953 0866 0082 0100 0227 0318 0509 0945 0866 2953 25 Solutions The commands shown for Exercise 31 but modi ed for 5 gtlt5 127 128 CHAPTER2 MatrixAlgebra b Let sk be the solution ofLsk tk for k 0 l 2 Then tk is the solution of Utk SM for k 0 1 2 The results are 100000 65556 65556 47407 153333 96667 118519 76667 s1 177500 t1 104444 s2 148889 t2 85926 187619 96667 153386 76667 171636 65556 124121 47407 47407 35988 35988 27922 92469 60556 72551 47778 53 120602 t3 69012 s4 96219 t4 54856 122610 60556 97210 47778 94222 35988 73104 27922 26 SOLUTIONS Notes This section is independent of Section 110 The material here makes a good backdrop for the series expansion of I39 1 because this formula is actually used in some practical economic work Exercise 8 gives an interpretation to entries of an inverse matrix that could be stated without the economic context 1 The answer to this exercise will depend upon the order in which the student chooses to list the sectors The important fact to remember is that each column is the unit consumption vector for the appropriate sector If we order the sectors manufacturing agriculture and services then the consumption matrix is 10 60 60 C 30 20 0 30 10 10 The intermediate demands created by the production vector x are given by CK Thus in this case the intermediate demand is 10 60 60 0 60 Cx 30 20 00 100 2 30 10 10 0 10 O 2 Solve the equation x Cx d for d x1 10 60 60 x1 9x1 6x2 6x3 0 dx Cx x2 30 20 00 x2 3x1 8x2 18 x3 30 10 10 2 3x1 1x2 9263 0 This system of equations has the augmented matrix 90 60 60 0 1 0 0 3333 30 80 00 18 N 0 1 0 3500 30 10 90 0 0 0 1 1500 so x 3333 35001500 26 Solutions 129 3 Solving as in Exercise 2 x1 10 60 60 x1 9x1 6x2 6x3 18 d x Cx x2 30 20 00 x2 3x1 8x2 0 x3 30 10 10 2 3x1 1x2 9x3 0 This system of equations has the augmented matrix 90 60 60 18 1 0 0 4000 30 80 00 0 N 0 1 0 1500 30 10 90 0 0 0 1 1500 so x 4000 15001500 4 Solving as in Exercise 2 x1 10 60 60 x1 9x1 6x2 6x3 18 dx Cx x2 30 20 00 x2 3x1 8x2 18 x3 30 10 10 2 3x1 1x2 9x3 0 This system of equations has the augmented matrix 90 60 60 18 1 0 0 7333 30 80 00 18 N 0 1 0 5000 30 10 90 0 0 0 1 3000 so x 7333 5000 3000 Note Exercises 241 may be used by students to discover the linearity of the Leontief model 1 1 5 1 50 16 1 50 110 5 x I C d 6 8 20 12 2 20 120 1 9 6 71 18 4021 3021 18 50 6 x I C d 5 8 11 2521 4521 11 45 7 a From Exercise 5 I CYI 16 1 12 2 I Cyld 16 1 1 16 x 1 1 12 2 0 12 which is the first column of I C 1 1 16 1 51 1116 b x2I C d2 12 2 30 1212 130 CHAPTER2 MatrixAlgebra 50 110 c From Exercise 5 the production x corressponding to d 20s x Note that d2 d d1 Thus x2 1 C 1d2 I C 1d d1 1 C 1d 1 C 1d1 x x1 8 21 Given 1 Cx d and I CAx Ad I CxAx I Cx I CAx d Ad Thus x Ax is the production level corresponding to a demand of d Ad b Since Ax I C 1Ad and Ad is the rst column of Ax will be the rst column of I C 9 In this case 8 2 0 I C 3 9 3 1 0 8 Row reduce I C d to nd 8 2 0 400 1 0 0 828 3 9 3 600 N 0 1 0 1310 1 0 8 800 0 0 1 1103 So x 828 1310 1103 10 From Exercise 8 the 139 j entry in I 7 C 1 corresponds to the effect on production of sector iwhen the nal demand for the output of sector j increases by one unit Since these entries are all positive an increase in the nal demand for any sector will cause the production of all sectors to increase Thus an increase in the demand for any sector will lead to an increase in the demand for all sectors H H Solution in study Guide Following the hint in the text compute pTx in two ways First take the transpose of both sides of the price equation p C T p v to obtain pT CTpvT CTpT vT pTCvT and rightmultiply by x to get pTx pTCvTxpTCxvTx Another way to compute pTx starts with the production equation x Cx d Left multiply by p7 to get pTxpTCxdpTCxpTd The two expression for pTx show that pTCxvTxpTCxpTd so vTx pT d The Study Guide also provides a slightly different solution 12 Since Dm11 CC2 Cm1 ICICCmICDm DW1 may be found iteratively by BMI I CDm 13 M The matrix I 7 C is 08412 00057 00264 03299 00089 01190 7 00063 08412 00057 00264 03299 00089 01190 7 00063 00064 00025 00304 00014 00083 07355 00436 00099 00083 00201 01506 06443 00139 00142 00070 00565 00495 06364 00204 00483 00081 00333 00295 06588 00237 00901 00996 01260 01722 07632 00126 00196 00098 00064 00132 so the augmented matrix I C d may be row reduced to nd 00064 00025 00304 00014 00083 07355 00436 00099 00083 00201 01506 06443 00139 00142 00070 00565 00495 06364 00204 00483 00081 00333 00295 06588 00237 00901 00996 0 1260 0 1722 07632 00126 00196 00098 00064 00132 1 0 0 0 0 0 0 99576 0 1 0 0 0 0 0 97703 0 0 1 0 0 0 0 51231 N 0 0 0 1 0 0 0 131570 0 0 0 0 1 0 0 49488 0 0 0 0 0 1 0 329554 0 0 0 0 0 0 1 13835 01594 03413 00236 00649 00020 03369 09988 01594 03413 00236 00649 00020 03369 09988 26 Solutions 74000 56000 10500 25000 17500 196000 5000 131 so x 99576 97703 51321 131570 49488 329554 13835 Since the entries in 1 seem to be accurate to the nearest thousand a more realistic answer would be x 100000 98000 51000 132000 49000 330000 14000 14 M The augmented matrix I C d in this case may be row reduced to nd 08412 00057 00264 03299 00089 01190 00063 00064 07355 01506 00565 00081 00901 00126 00025 00436 06443 00495 00333 00996 00196 00304 00099 00139 06364 00295 01260 00098 00014 00083 00 142 00204 06588 01722 00064 00083 00201 00070 00483 00237 07632 00132 0 1594 03413 00236 00649 00020 03369 09988 99640 75548 14444 33501 23527 263985 6526 132 CHAPTER2 MatrixAlgebra 0 134034 0 131687 0 69472 0 176912 0 66596 0 443773 0 0 0 0 1 18431 so x 134034 131687 69472 176912 66596 443773 18431 To the nearest thousand x 134000 132000 69000 177000 67000 444000 18000 Z OOOOOOD I OOOOp IO OOOp IOO OOp IOOO OOp IOOOO HOOOOO 15 M Here are the iterations rounded to the nearest tenth Kw 740000 560000 105000 250000 175000 1960000 50000 x0 893442 777305 267081 723347 303256 2651582 93278 X2 946812 877145 375773 1005205 385980 2965638 114800 X3 970919 925731 4386781154570 434910 3123190125988 X4 982916 950332 473145 1232025 462470 3205024 131855 5 989072 963053 491606 1272137 477564 324796 134938 X6 992266 969696 5013961292967 485693 3270538 136559 x7 993931 973178 506564 1303816 490028 3282409 137411 x8 994800 975007 509287 1309480 492325 3288647 137859 x9 995255 975968 510719 131244 493538 3291923 138094 x00 995494 976472 511472 1313992 494177 3293644 138217 xm 995619 976737 511868 1314804 494513 3294547 138282 xuz 995684 976876 512075 1315230 494690 329502 138316 so Km is the rst vector whose entries are accurate to the nearest thousand The calculation of xuz takes about 1260 ops while the row reduction above takes about 550 ops If C is larger than 20 x 20 then fewer ops are required to compute xuz by iteration than by row reduction The advantage of the iterative method increases with the size of C The matrix C also becomes more sparse for larger models so fewer iterations are needed for good accuracy 27 SOLUTIONS Notes The content of this section seems to have universal appeal with students It also provides practice with composition of linear transformations The case study for Chapter 2 concerns computer graphics 7 see this case study available as a project on the website for more examples of computer graphics in action The Study Guide encourages the student to examine the book by Foley referenced in the text This section could form the beginning of an independent study on computer graphics with an interested student 27 Solutions 133 1 Refer to Example 5 The representation in homogenous coordinates can be written as a partitioned matrix A 0 of the form of I where A is the matrix of the linear transformation Since in this case 1 25 0 I the representatlon of the transformatlon w1th respect to homogenous coordmates 1s 25 1 A 0 Note The Study Guide shows the student why the action of 0T 1 on the vector corresponds to the action ofA on x 1 0 2 The matr1x of the transformatlon 1s A 0 1 so the transformed data matr1x 1s 1 0 5 2 4 5 2 4 AD 01L 2 3i 0 2 3 Both the original triangle and the transformed triangle are shown in the following sketch 3 Following Examples 4 22 22 0 1 0 3 22 22 2 22 22 0 0 1 1 22 22 2 2 0 0 l 0 0 l 0 0 l 28 0 0 l 0 2 8 0 l6 4 0 12 0 0 l 3 70 0 l 0 0 l 0 0 l o t N E Ox 7 2 12 0 1 0 0 52 12 0 12 52 0 0 1 0 12 32 0 0 0 10 0 1 0 0 1 El39 1 o o IE2 12 0 IE2 12 0 6 0 1 0 12 IE2 0 12 2 0 0 0 1 0 0 1 0 0 1 134 CHAPTER2 gt0 p A 6 Matrix Algebra A 600 rotation about the origin is given in homogeneous coordinates by the matrix 1 2 3 2 0 J3 2 1 2 0 To rotate about the point 6 8 first translate by 78 then rotate about the 0 0 1 origin then translate back by 6 8 see the Practice Problem in this section A 600 rotation about 6 8 is thus is given in homogeneous coordinates by the matrix 1 0 6 12 2 0 1 0 6 12 2 34J 0 1 8 J 2 12 0 0 1 8 z 12 4 36 0 0 1 0 0 1 0 0 1 0 0 1 A 450 rotation about the origin is given in homogeneous coordinates by the matrix J5 2 45 2 0 J2 2 J2 2 0 To rotate about the point 3 7 rst translate by 73 77 then rotate about the 0 0 l origin then translate back by 3 7 see the Practice Problem in this section A 450 rotation about 3 7 is thus is given in homogeneous coordinates by the matrix 0 3 52 452 010 3 52 1 452 32J o 1 7 452 52 0 o 1 7J 2 Jiz 7 5J5 0 0 1 0 0 1 0 0 1 0 0 1 To produce each entry in BD two multiplications are necessary Since ED is a 2 X 200 matrix it will take 2X 2X 200 800 multiplications to compute BD By the same reasoning it will take 2X 2X 200 800 multiplications to computeABD Thus to computeABD from the beginning will take 800 800 1600 multiplications To compute the 2X 2 matrix AB it will take 2 X 2 X 2 8 multiplications and to compute ABD it will take 2X 2X 200 800 multiplications Thus to compute ABD from the beginning will take 8 800 808 multiplications For computer graphics calculations that require applying multiple transformations to data matrices it is thus more efficient to compute the product of the transformation matrices before applying the result to the data matrix Let the transformation matrices in homogeneous coordinates for the dilation rotation and translation be called respectively D and R and T Then for some value of s 0 h and k s 0 0 cosgo sing0 0 1 0 h D 0 s 0 R singp cosgp 0 T 0 1 k 0 0 1 0 0 l 0 0 1 Compute the products of these matrices 3 cos 0 s singp 0 3 cos 0 s sin 0 0 DR ssingp scosgp 0 RD ssingp scosgp 0 0 0 1 0 0 1 27 Solutions 135 s 0 sh s 0 h DTOSskTDOSk 0 0 l 0 0 1 cosgo sing0 hcosgo ksingo cosgo sing0 h RT singo cosgo hsing0kcosg0 TR singo cosgp k 0 0 1 0 0 1 Since DR RD DT 7 TD and RT 7 TR D and R commute D and T do not commute and R and T do not commute 11 To simplifyA 2A 1 completely the following trigonometric identities will be needed 1 tanq0cosg0 iscosg0 sing0 2 secgo tan qosin 0 r stingp 1 15 cos 0 Using these identities secgo tan 0 0 1 0 0 A2A1 0 l 0 singp cosgp 0 0 0 1 0 0 l secqo tangosingo tang0cosg0 0 sin 0 cos 0 0 0 0 1 cos 0 sin 0 0 sin 0 cos 0 0 0 0 1 which is the transformation matrix in homogeneous coordinates for a rotation in R2 12 To simplify this product completely the following trigonometric identity will be needed l cosgo singo singo 1cosgp tango2 This identity has two important consequences l tangJ2singz1l 1c sing0 cosgp smgo cosq0 tang02 tang02 cosg0ltang02 cosg0l1 8mg sing0 co go The product may be computed and simplified using these results 1 tangp2 0 l 0 0 l tangp2 0 1 0 0 l 0 singp 0 l 0 0 0 1 0 0 l 0 0 1 l tanng2sing0 tang02 0 l tang02 0 singp l 0 0 1 0 0 0 1 0 0 1 136 CHAPTER 2 543 Matrix Algebra cosgp tangp2 0 1 tangp2 0 singp 1 0 0 1 0 0 0 1 0 0 1 cos 0 cos g0 tango2 tan 0 2 0 singp sinq0tang021 0 0 0 1 cosgp sing0 0 singp cosgp 0 0 0 1 which is the transformation matrix in homogeneous coordinates for a rotation in R2 Consider rst applying the linear transformation on R2 whose matrix is A then applying a translation by the vector p to the result The matrix quot 1n 39 quot of the linear 1 of the A 0 transformation is of I while the matrix 1 quot in39 quot I translation is of Applying these transformations in order leads to a transformation whose matrix representation in homogeneous coordinates is I p A 0 A p 0T 1 0T 1 0T 1 which is the desired matrix The matrix for the transformation in Exercise 7 was found to be 1 2 32 3 N 2 12 4 3J 0 0 1 A p Th1s matrlx 1s of the form of 1 where 12 32 34J A p JE2 12 4 3 By Exercise 13 this matrix may be written as L Elquot 1 that is the composition of a linear transformation on R2 and a translation The matrixA is the matrix of a rotation about the origin in R2 Thus the transformation in Exercise 7 is the composition of a rotation 34J 4 3J 39 about the origin and a translation by p 15 16 p A l p A on 27 Solutions 137 Since X Y Z H the corresponding point in R3 has coordinates l xyz Li 4 H H H 2 14 2 14 The homogeneous coordinates 1 72 3 4 represent the point 14 24 34 14 12 34 while the homogeneous coordinates 10 720 30 40 represent the point 1040 2040 3040 1 4 12 34 so the two sets of homogeneous coordinates represent the same point in R3 H j 12 63 glp loop Follow Example 7a by first constructing that 3 X 3 matrix for this rotation The vector e1 is not changed by this rotation The vector e2 is rotated 60 toward the positive Zaxis ending up at the point 0 cos 60 sin 60 0 1 2 J3 2 The vector 63 is rotated 60 toward the negative yaxis stopping at the point 0 cos 150 sin 150 0 32 12 The matrixA for this rotation is thus 1 0 0 A 0 1 2 J 2 0 J 2 1 2 so in homogeneous coordinates the transformation is represented by the matrix 1 0 0 0 A 0 0 12 J 2 0 OT lie 2 12 0 0 0 0 1 First construct the 3 X 3 matrix for the rotation The vector e1 is rotated 30 toward the negative yaxis ending up at the point cos730 sin 730 0 J3 2 1 2 0 The vector e2 is rotated 60 toward the positive xaxis ending up at the point cos 60 sin 60 0 1 2 J3 2 0 The vector e3 is not changed by the rotation The matrixA for the rotation is thus 52 12 0 A 12 J 2 0 0 0 1 so in homogeneous coordinates the rotation is represented by the matrix 52 12 0 0 A 0 12 JE2 0 0 0T 1 0 0 1 0 0 0 0 1 Following Example 7b in homogeneous coordinates the translation by the vector 5 72 1 is represented by the matrix 1 0 5 2 Op IOO 0 1 0 0 0 0 138 p A D N G CHAPTER2 MatrixAlgebra Thus the complete transformation is represented in homogeneous coordinates by the matrix 100 5J 2 1200 JEz 120 5 0102 12J 200 12J 20 2 0011 0 010 0 011 00010 001 0 00 Referring to the material preceding Example 8 in the text we find that the matrixP that performs a perspective projection with center of projection 0 0 10 is 1000 0100 0000 700 11 The homogeneous coordinates of the vertices of the triangle may be written as 42 12 4 1 6 4 2 1 and 2 2 6 1 so the data matrix for S is 742 6 2 12 4 2 4 2 6 7 1 1 1 and the data matrix for the transformed triangle is 7 0 0 0 42 6 2 42 6 2 010012421242 0 0 0 0 4 2 6 0 0 0 70 0 1 1 1 1 1 6 8 4 Finally the columns of this matrix may be converted from homogeneous coordinates by dividing by the nal coordinate 4212 0 6 gt 426 126 06 7 2 0 6 40 8 gt6828087550 2 2 0 4 gt 24 24 04 5 5 0 So the coordinates of the vertices of the transformed triangle are 7 2 0 75 5 0 and 5 5 0 As in the previous exercise the matrix P that performs the perspective projection is 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 The homogeneous coordinates of the vertices of the triangle may be written as 9 3 75 1 12 8 2 1 and 18 27 1 1 so the data matrix for S is 9 12 18 3 8 27 5 2 1 1 1 1 28 Solutions 139 and the data matrix for the transformed triangle is l 0 0 0 9 12 18 9 12 18 0 l 0 0 3 8 27 3 8 27 0 0 0 0 5 2 1 0 0 0 0 0 l l l l l 15 8 9 Finally the columns of this matrix may be converted from homogeneous coordinates by dividing by the nal coordinate 9 3 015 gt915 315 015 6 2 0 12 8 0 8 gt 128 88 0815 10 0 18 27 0 9 gt 189 279 09 2 3 0 So the coordinates of the vertices of the transformed triangle are 6 2 0 15 10 0 and 2 3 0 21 M Solve the given equation for the vector R G B giving R 61 29 15 71 X 22586 10395 3473 X G 35 59 063 Y 13495 23441 0696 Y B 04 12 787 Z 0910 3046 12777 Z 22 M Solve the given equation for the vector R G B giving R 299 587 114 71 Y 10031 9548 6179 Y G 596 275 321 I 9968 2707 6448 B 212 528 311 Q 10085 11105 16996 Q N 28 SOLUTIONS Notes Cover this section only if you plan to skip most or all of Chapter 4 This section and the next cover everything you need from Sections 4 1416 to discuss the topics in Section 49 and Chapters 577 except for the general inner product spaces in Sections 67 and 68 Students may use Section 42 for review particu larly the Table near the end of the section The nal subsection on linear transformations should be omitted Example 6 and the associated exercises are critical for work with eigenspaces in Chapters 5 and 7 Exercises 3136 review the Invertible Matrix Theorem New statements will be added to this theorem in Section 29 Key Exercises 5720 and 23726 1 The set is closed under sums but not under multiplication u by a negative scalar A counterexample to the subspace condition is shown at the right 1u Note Most students prefer to give a geometric counterexample but some may choose an algebraic calcu lation The four exercises here should help students develop an understanding of subspaces but they may be insuf cient if you want students to be able to analyze an unfamiliar set on an exam Developing that skill seems more appropriate for classes covering Sections 41746 140 CHAPTER2 MatrixAlgebra 2 The set is closed under scalar multiples but not sums For example the sum of the vectors u and v shown here is not in H j 3 No The set is not closed under sums or scalar multiples The subset ll 11 V consisting of the points on the line x2 x1 is a subspace so any 3n n V l counterexample must use at least one point not on this line Here are two counterexamples to the subspace conditions v 4 No The set is closed under sums but not under multiplication by a negative scalar 71 5 The vector w is in the subspace generated by v1 and v2 if and only if the vector equation xlvl xzvz w is consistent The row operations below show that w is not in the subspace generated by v1 and v2 2 4 8 2 4 8 4 8 vlvzw 3 5 201 100 9 10 5 8 9 0 2 1100 6 The vector u is in the subspace generated by v1 v2 v3 if and only if the vector equation xlvl xzvz x3v3 u is consistent The row operations below show that u is not in the subspace generated by V1V2V3 1 4 5 4 1 4 5 4 D 4 5 4 2 7 81001220 22 Vlv2v3uw4 9 6 7N0 7 14 9N000 3 7 5 5 0 5 10 7 0 0 0 17 Note For a quiz you could use w l 73 ll 8 which is in Spanv1 v2 v3 7 21 There are three vectors v1 v2 and v3 in the set v1 v2 v3 b There are infmitely many vectors in Spanv1 v2 v3 ColA c Deciding whether p is in Col A requires calculation 2 3 4 6 2 3 4 6 3 4 6 A pN 8 8 6 100 4 10140 1014 6 7 7 11 0 2 5 7 0 0 0 0 The equationAx p has a solution so p is in ColA 28 Solutions 141 3 2 0 1 3 2 0 1 2 0 1 A p 0 2 6 14 N 0 2 6 14 N 0 6 14 6 3 3 9 0 1 3 7 0 0 0 0 9 Yes the augmented matrix A p corresponds to a consistent system so p is in ColA To determine whether p is in NulA simply compute Ap UsingA and p as in Exercise 7 2 3 4 6 2 Ap 8 8 6 10 62 SinceAp 0 p is notin NulA 6 7 7 11 29 D 10 To determine whether 11 is in NulA simply compute Au UsingA as in Exercise 7 and u 72 3 1 3 2 0 2 0 Au 0 2 6 3 0Yesuis111Nu1A 6 3 3 1 0 11 p 4 and q 3 NulA is a subspace of R4 because solutions of Ax 0 must have 4 entries to match the columns ofA ColA is a subspace of R3 because each column vector has 3 entries 12 p 3 and q 4 NulA is a subspace of R3 because solutions ofo 0 must have 3 entries to match the columns of A ColA is a subspace of R4 because each column vector has 4 entries 13 To produce a vector in ColA select any column ofA For Nul A solve the equationAx 0 Include an augmented column of zeros to avoid errors 3 2 1 5 0 3 2 1 5 0 3 2 1 5 9 4170024 80024 80 9 2 5 1 0 0 4 8 16 0 0 0 0 0 0 3 21 5 0 0 1 10 x3x40 012 4 00 2 4 0 2x3 4x40 0 0 0 0 0 0 0 0 0 0 00 The general solution is x1 x3 7 x4 and x2 72x3 4x4 with x3 and 26 free The general solution in parametric vector form is not needed All that is required here is one nonzero vector So choose any values for x3 and 26 not both zero For instance set x3 l and x4 0 to obtain the vector 1 72 1 0 in Nul A Note Section 28 of Study Guide introduces the ref command or rref depending on the technology which produces the reduced echelon form of a matrix This will greatly speed up homework for students who have a matrix program available 14 To produce a vector in ColA select any column of A For Nul A solve the equationAx 0 1 2 3 0 1 2 3 0 1 2 3 0 0 13 0 4 570 0 3 5 01530 0 530 N 0 0 0 5 1 0 0 0 9 15 0 0 0 0 0 0 0 0 2 7 11 0 0 3 5 0 0 0 0 0 0 0 0 The general solution is x1 l3x3 and x 753 203 with x3 free The general solution in parametric vector form is not needed All that is required here is one nonzero vector So choose any values of 03 and 26 not both zero For instance set x3 3 to obtain the vector 1 75 3 in NulA 142 Equot 1 l CHAPTER 2 Matrix Algebra Yes LetA be the matrix whose columns are the vectors given ThenA is invertible because its determinant is nonzero and so its columns form a basis for R2 by the Invertible Matrix Theorem or by Example 5 Other reasons for the invertibility of A could be given No One vector is a multiple of the other so they are linearly dependent and hence cannot be a basis for any subspace No Place the three vectors into a 3X3 matrixA and determine whetherA is invertible 0 5 6 1 7 3 1 7 3 7 3 A1 7 30 560 5 60 6 2 4 5 2 4 5 0 10110 0 The matrixA has three pivots so A is invertible by the IMT and its columns form a basis for R3 as pointed out in Example 5 Yes Place the three vectors into a 3X3 matrixA and determine whetherA is invertible 1 5 71 5 7 5 7 A1 100 4 70 7 2 2 5 0 8 9 0 063 The matrixA has three pivots so A is invertible by the IMT and its columns form a basis for R3 as pointed out in Example 5 No The vectors cannot be a basis for R3 because they only span a plan in R3 Or point out that the l 5 columns of the matrix 1 1 cannot possibly span R3 because the matrix cannot have a pivot in 2 2 every row So the columns are not a basis for R3 Note The Study Guide warns students not to say that the two vectors here are a basis for R2 No The vectors are linearly dependent because there are more vectors in the set than entries in each vector Theorem 8 in Section 17 So the vectors cannot be a basis for any subspace False See the de nition at the beginning of the section The critical phrases for each are missing True See the paragraph before Example 4 False See Theorem 12 The null space is a subspace of Rquot not Rm True See Example 5 3 59 True See the rst part of the solution of Example 8 F False See the de nition at the beginning of the section The condition about the zero vector is only one of the conditions for a subspace True See Example 3 True See Theorem 12 False See the paragraph after Example 4 False See the Warning that follows Theorem 13 supn 28 Solutions 143 4 5 9 2 2 6 5 23 Solution in Study Guide A 6 5 1 12 N 0 CD 5 6 The echelon form identi es 3 4 8 3 0 0 0 0 4 5 columns 1 and 2 as the pivot columns A basis for ColA uses columns 1 and 2 of A 6 5 This is not 3 4 the only choice but it is the standard choice A wrong choice is to select columns 1 and 2 of the echelon form These columns have zero in the third entry and could not possibly generate the columns displayed in A 24 For NulA obtain the reduced and augmented echelon form forAx 0 0 4 7 0 4x3 7x4 0 0 D 5 6 0 This corresponds to 5x3 6x4 0 0 0 0 0 0 0 0 Solve for the basic variables and write the solution ofo 0 in parametric vector form x1 4 7x4 4 7 4 7 x2 5x3 6x4 5 6 x3 x4 Bas1s for NulA x3 x3 1 0 x4 x4 0 l 0 Notes 1 A basis is a set of vectors For simplicity the answers here and in the text list the vectors without enclosing the list inside set brackets This style is also easier for students I am careful however to distinguish between a matrix and the set or list whose elements are the columns of the matrix 2 Recall from Chapter 1 that students are encouraged to use the augmented matrix when solvingAx 0 to avoid the common error of misinterpreting the reduced echelon form of A as itself the augmented matrix for a nonhomogeneous system 3 Because the concept of a basis is just being introduced I insist that my students write the parametric vector form of the solution of Ax 0 They see how the basis vectors span the solution space and are obviously linearly independent A shortcut which some instructors might introduce later in the course is only to solve for the basic variables and to produce each basis vector one at a time Namely set all free variables equal to zero except for one free variable and set that variable equal to a suitable nonzero number 3 9 2 7 3 6 9 3 2 24 A 2 6 4 8 N 0 0 GD 5 BasisforColA 2 4 3 9 2 2 0 0 0 0 3 2 For Nul A obtain the reduced and augmented echelon form forAx 0 3 0 150 0 3x2 150x4 0 0 0 D 125 0 This corresponds to 125x4 0 0 0 0 0 0 0 0 Solve for the basic variables and write the solution of AX 0 in parametric vector form x1 3x2 15x4 3 15 3 15 x2 x2 1 0 0 x2 x4 Bas1s for NulA x3 125x4 0 125 0 125 x4 x4 0 1 0 1 144 CHAPTER2 MatrixAlgebra 1 4 8 3 7 4 8 0 5 1 4 3 1 2 7 3 0 5 0 1 1 2 3 25 A N Basis for ColA 2 9 5 5 0 0 0 CD 4 2 2 5 3 6 9 5 2 0 0 0 0 0 3 6 5 For Nul A obtain the reduced and augmented echelon form forAx 0 D 0 2 0 7 0 2203 7x5 0 0 25 0 5 0 25 5 0 A 0 N D x3 x5 0 0 0 CD 4 0 4x5 0 0 0 0 0 0 0 0 0 x1 22 7x5 2 7 x2 25x3 5265 25 5 Thesolution of Ax Oin parametric vector form x3 x3 x3 1 x5 0 x4 4x5 0 4 x5 x5 0 l u v Basis for NulA u v Note The solution above illustrates how students could write a solution on an exam when time is precious namely describe the basis by giving names to appropriate vectors found in the calculations 3 1 7 3 9 1 7 0 6 3 1 3 2 2 2 7 5 0 4 0 3 2 7 26 N Bas1s for ColA 5 9 3 4 0 0 0 D 1 5 9 3 2 6 3 7 0 0 0 0 0 2 6 3 For NulA D 0 3 0 25 0 3x 25265 0 A 0 0 D 2 0 15 0 2203 15265 0 0 0 0 1 0 x5 0 0 0 0 0 0 0 0 0 The solution of Ax 0 in parametric vector form x1 3g 25x5 3 25 x2 2g l5x5 2 l5 x3 9 263 1 265 0 x4 x5 0 l Basis for NulA u v 1 x5 x5 0 u v 27 Construct a nonzero 3gtlt3 matriXA and construct b to be almost any convenient linear combination of the columns of A DJ p A 43 UI 43 Ch 43 l Since ColA is the set of all linear combinations of a1 28 Solutions 145 The easiest construction is to write a 3 gtlt3 matrix in echelon form that has only 2 pivots and let b be any vector in R3 whose third entry is nonzero Solution in Study Guide A simple construction is to write any nonzero 3gtlt3 matrix whose columns are obviously linearly dependent and then make b a vector of weights from a linear dependence relation among the columns For instance if the first two columns ofA are equal then b could be 1 fl 0 ap the set 211 ap spans ColA Because 211 ap 1s also linearly independent it is a basis for ColA There is no need to discuss pivot columns and Theorem 13 though a proof could be given using this information If Col F R5 then the columns of F do not span R5 Since F is square the IMT shows thatF is not invertible and the equation F x 0 has a nontrivial solution That is Nul F contains a nonzero vector Another way to describe this is to write Nul F 0 If Nul R contains nonzero vectors then the equation Rx 0 has nontrivial solutions Since R is square the IMT shows thatR is not invertible and the columns of R do not span R6 So Col R is a subspace of R6 but Col R R6 If Col Q R4 then the columns of Q span R4 Since Q is square the IMT shows that Q is invertible and the equation Qx b has a solution for each h in R4 Also each solution is unique by Theorem 5 in Section 22 If Nul P 0 then the equation PX 0 has only the trivial solution Since P is square the IMT shows that P is invertible and the equation PX b has a solution for each h in R5 Also each solution is unique by Theorem 5 in Section 22 If the columns of B are linearly independent then the equation Bx 0 has only the trivial zero solution That is Nul B 0 Ifthe columns ofA form a basis they are linearly independent This means thatA cannot have more columns than rows Since the columns also span R A must have a pivot in each row which means that A cannot have more rows than columns As a result A must be a square matrix M Use the command that produces the reduced echelon form in one step ref or rref depending on the program See the Section 28 in the Study Guide for details By Theorem 13 the pivot columns of A form a basis for ColA 3 5 0 1 3 0 25 45 35 3 5 7 9 4 9 11 0 15 25 15 7 9 A N Basis for ColA 5 7 2 5 7 0 0 0 0 5 7 3 7 3 4 0 0 0 0 0 0 3 7 For NulA obtain the solution of Ax 0 in parametric vector form 25x3 45x4 35x5 0 15x3 25x4 15x5 0 25x3 45x4 35x5 l5x3 25x4 15x5 x3 x4 and x5 are free x1 Solution x2 146 CHAPTER2 MatrixAlgebra x1 25g 45x4 35x5 25 45 35 x2 15g 25x4 15x5 15 25 15 X x3 x3 x3 1 x4 0 x5 0 Jgux4vx5w x4 x4 0 l 0 x5 x5 0 0 1 By the argument in Example 6 a basis for NulA is u v w 520 8 8 00 60122 4 1 2 8 9 0 D 0 154 309 38 M A N 51351900 47 94 8 5 6 8 5 0 0 0 0 0 5 0 4 2 Thep1vot columns ofA formabas1s for ColA 5 1 3 8 5 6 60x4 122x5 0 For NulA solveAx0 l54x4 309x5 0 47x4 94x5 0 x1 60x4 l22x5 154 309 Solution x2 x4 x5 47x4 94x5 x4 and x5 are free x1 60x4 122x5 60 122 x2 154x4 309x5 154 309 X x3 47x494x5 x4 47 965 94 x4ux5v x4 x4 1 0 x5 x5 0 1 By the method of Example 6 a basis for NulA is u v Note The Study Guide for Section 28 gives directions for students to construct a review sheet for the concept of a subspace and the two main types of subspaces Col A and Nul A and a review sheet for the concept of a basis I encourage you to consider making this an assignment for your class 29 SOLUTIONS Notes This section contains the ideas from Sections 44 46 that are needed for later work in Chapters 577 If you have time you can enrich the geometric content of coordinate systems by discussing crystal lattices Example 3 and Exercises 35 and 36 in Section 44 Some students might pro t from reading Examples 173 from Section 44 and Examples 2 4 and 5 from Section 46 Section 45 is probably not a good reference for students who have not considered general vector spaces Coordinate vectors are important mainly to give an intuitive and geometric feeling for the isomorphism between a k dimensional subspace and Rk If you plan to omit Sections 54 56 57 and 72 you can safely omit Exercises 178 here 29 Solutions 147 Exercises 1716 may be assigned after students have read as far as Example 2 Exercises 19 and 20 use the Rank Theorem but they can also be assigned before the Rank Theorem is discussed The Rank Theorem in this section omits the nontrivial fact about RowA which is included in the Rank Theorem of Section 46 but that is used only in Section 74 The row space itself can be introduced in Section 62 for use in Chapter 6 and Section 74 Exercises 9716 include important review of techniques taught in Section 28 and in Sections 12 and 25 They make good test questions because they require little arithmetic My students need the practice here Nearly every time I teach the course and start Chapter 5 I find that at least one or two students cannot find a basis for a twodimensional eigenspace 3 1 If xB 2 then x is formed from b1 and b2 using weights 3 and 2 x3b12b2 ii i iizili 1 2 If x3 3 then x is formed from b1 and b2 using weights 71 and 3 2 3 11 x71b13b2 1 J3MLl 3 To find cl and C that satisfy x cl b1 02b2 row reduce the augmented matrix 1 2 3 1 2 3 1 0 7 X N N f 0116 can er e ama 1X e ua 1011 as b1 b2 o t tr q t 4 7 7 0 1 5 0 1 5 suggested by Exercise 7 and solve using the matrix inverse In either case cl 7 x 3 c2 5 1 3 7 1 3 7 1 0 5 4 As in Exerc1se 3 b1 b2 x N N 5 and 3 5 5 0 4 16 0 1 4 cl 5 x B CZ 4 1 3 4 1 3 4 1 0 14 cl 14 5b1b2 x 5 7 10 N 0 8 10 N 0 1 54 x3 02 54 3 5 7 0 4 5 0 0 0 148 CHAPTER2 MatrixAlgebra 73 7 11 1 5 0 1 0 752 6 b1 b2 x 1 5 0 N 0 22 11 N 0 1 12 74 76 7 0 14 7 0 0 0 cl 752 x 3 c2 12 7 Fig 1 suggests that w 2b1 7 b2 and x l5b1 5b2 in which case 2 15 w3 1 and XB 5 To con rm x3 compute 3 1 15b15b215 5 0 2 Figure 1 Figure 2 Note Figures 1 and 2 display what Section 44 calls Bgraph paper 8 Fig 2 suggests that x 2b1 7 b2 y l5b1 b2 and z 7b1 7 5b2 If so then xs y3 and z3 To confirm y3 and z3 compute 15b1b2151311111i1 and7539 71131751311217 1 3 2 4 3 2 4 73 9 71 5 0 0 7 9 The information A N 6 is enough to see that columns 1 3 and 4 of 2 76 4 3 0 0 0 9 7412270000 lt 1 2 74 71 5 A form a bas1s for ColA 4 73 74 2 7 Columns 1 2 and 4 of the echelon form certain cannot span ColA since those vectors all have zero in their fourth entries For NulA use the reduced echelon form augmented with a zero column to insure that the equationAx 0 is kept in mind p n 29 Solutions 149 D 3 0 0 0 3x2 0 x1 3x2 3 3 0 0 0 0 0 1 1 x x2 x2 x2 So is 0 0 0 D 0 0 2g 0 0 0 0 0 0 0 0 x2 is the free variable x4 0 0 0 a basis for Nul A From this information dim C01 A 3 becauseA has three pivot columns and dim Nul A 1 because the equationAx 0 has only one free variable 2 9 5 4 2 9 5 4 1 6 5 3 0 D 0 7 The mformatlonA N shows that columns 1 2 0 6 1 2 0 0 0 2 4 1 9 1 9 0 0 0 0 0 1 2 5 1 1 5 and 4 ofA form a bas1s for ColA 2 1 For NulA 4 l 1 D 0 3 0 0 0 3x3 2 0 3 0 7 0 3 7 0 A N G I x3 x5 0 0 0 2 0 2x5 2 0 0 0 0 0 0 0 x3 and x5 are free variables x1 3x3 3 0 3 0 x2 32 7x5 3 7 3 7 x x3 2 263 lx5 0 BasisforNulA 1 0 x4 2x5 0 2 0 2 x5 x5 0 l 0 1 From this dim ColA 3 and dim NulA 2 1 2 5 0 1 D 2 5 0 1 2 5 8 4 3 0 2 4 5 The information A N shows that columns 1 2 3 9 9 7 2 0 0 0 D 2 3 10 7 11 7 0 0 0 0 0 1 2 0 2 5 4 and 4 ofA form a bas1s for ColA 3 9 7 For NulA 3 10 11 0 9 0 5 0 9x3 5x5 2 0 0 2 0 3 0 2x 3x 0 A N D I 3 5 0 0 0 D 2 0 2x5 0 0 0 0 0 0 0 x3 and x5 are free variables 150 CHAPTER2 MatrixAlgebra x1 9x3 5x5 9 5 9 5 x2 2x3 3x5 2 3 2 3 x x3 g x3 1 x5 0 Basis forNulA 1 0 x4 2x5 0 2 0 2 x5 x5 0 l 0 1 From this dim ColA 3 and dim NulA 2 1 2 4 3 2 4 3 3 5 10 9 7 8 0 2 0 12 The information A N D shows that columns 1 3 4 8 9 2 7 0 0 0 2 4 5 0 6 0 0 0 0 0 1 4 3 5 9 8 and 5 ofA form a bas1s for ColA 4 9 7 For NulA 2 5 6 D 2 0 5 0 0 2x2 5x4 0 0 0 2 0 0 2 0 A 0 N x4 0 0 0 0 D 0 0 0 0 0 0 0 0 x2 and x4 are free variables x1 2x2 5x4 7 2 57 2 5 x2 x2 1 0 1 0 x x3 2x4 x2 0 x4 2 BasisforNulA 0 2 x4 x4 0 l 0 1 x5 0 7 0 0 0 0 From this dim ColA 3 and dim NulA 2 13 The four vectors span the column space H of a matrix that can be reduced to echelon form 1 32 4 71 32 471 32 4 32 4 39 15 005 7 005 7 00 7 0 0050005000 4122770010 9700050000 Columns 1 3 and 4 of the original matrix form a basis for H so dim H 3 Note Either Exercise 13 or 14 should be assigned because there are always one or two students who confuse ColA with Nul A Or they wrongly connect set of linear combinations with parametric vector form of the general solution of Ax 0 14 The ve vectors span the column space H of a matrix that can be reduced to echelon form 1 2 0 1 3 1 2 0 1 3 2 0 1 3 1 3 2 4 8 0 1 2 3 5 0 ED 2 3 5 2 1 6 7 9 N 0 3 6 9 15 N 0 0 0 0 0 5 6 8 7 5 0 4 8 12 20 0 0 0 0 Columns 1 and 2 of the original matrix form a basis for H so dim H 2 19 29 Solutions 151 ColA R3 becauseA has a pivot in each row and so the columns ofA span R3 NulA cannot equal R2 because NulA is a subspace of R5 It is true however that NulA is twodimensional Reason the equationAx 0 has two free variables becauseA has ve columns and only three of them are pivot columns ColA cannot be R3 because the columns ofA have four entries In fact ColA is a 3dimensional subspace of R4 because the 3 pivot columns of A form a basis for ColA SinceA has 7 columns and 3 pivot columns the equationAx 0 has 4 free variables So dim NulA 4 a True This is the de nition of a B coordinate vector 6quot False Dimension is de ned only for a subspace A line must be through the origin in Rquot to be a subspace of Rquot P True The sentence before Example 1 concludes that the number of pivot columns of A is the rank of A which is the dimension of ColA by de nition True This is equivalent to the Rank Theorem because rankA is the dimension of ColA FL 5quot True by the Basis Theorem In this case the spanning set is automatically a linearly independent set F True This fact is justi ed in the second paragraph of this section 9 True See the second paragraph after Fig 1 P False The dimension of NulA is the number of free variables in the equationAx 0 See Example 2 P True by the de nition of rank D True by the Basis Theorem In this case the linearly independent set is automatically a spanning set The fact that the solution space of Ax 0 has a basis of three vectors means that dim Nul A 3 Since a 5gtlt7 matrixA has 7 columns the Rank Theorem shows that rankA 7 7 dim NulA 4 Note One can solve Exercises 19722 without explicit reference to the Rank Theorem For instance in Exercise 19 if the null space of a matrix A is threedimensional then the equation AX 0 has three free variables and three of the columns of A are nonpivot columns Since a 5gtlt7 matrix has seven columns A must have four pivot columns which form a basis of ColA So rankA dim ColA 4 20 N p A N N A 4gtlt5 matrixA has 5 columns By the Rank Theorem rankA 5 7 dim NulA Since the null space is threedimensional rankA 2 A 7gtlt6 matrix has 6 columns By the Rank Theorem dim NulA 6 7 rankA Since the rank is four dim NulA 2 That is the dimension of the solution space of Ax 0 is two The wording of this problem was poor in the rst printing because the phrase it spans a four dimensional subspace was never de ned Here is a revision thatI will put in later printings of the third edition Show that a set v1 v5 4 Solution Suppose that the subspace H Spanv1 v5 is fourdimensional If v1 v5 were linearly independent it would be a basis for H This is impossible by the statement just before the de nition of dimension in Section 29 which essentially says that every basis of a pdimensional subspace consists of p vectors Thus v1 v5 must be linearly dependent v5 in Rquot is linearly dependent ifdim Spanv1 A 3gtlt4 matrixA with a twodimensional column space has two pivot columns The remaining two columns will correspond to free variables in the equationAx 0 So the desired construction is possible 152 CHAPTER 2 Matrix Algebra I gt1lt gt1lt gt1lt There are six possible locations for the two pivot columns one of which is 0 I A simple 0 0 0 0 construction is to take two vectors in R3 that are obviously not linearly dependent and put two copies of these two vectors in any order The resulting matrix will obviously have a twodimensional column space There is no need to worry about whether NulA has the correct dimension since this is guaranteed by the Rank Theorem dim NulA 4 7 rankA A rank 1 matrix has a onedimensional column space Every column is a multiple of some xed vector To construct a 4gtlt3 matrix choose any nonzero vector in R4 and use it for one column Choose any multiples of the vector for the other two columns 25 The p columns of A span ColA by de nition If dim ColA p then the spanning set of p columns is N gt0 9 gt39 automatically a basis for ColA by the Basis Theorem In particular the columns are linearly independent If columns a1 a3 a5 and 36 of A are linearly independent and if dim ColA 4 then 211 a3 a5 36 is a linearly independent set in a 4dimensional column space By the Basis Theorem this set of four vectors is a basis for the column space a Start withB b1 hp andA a1 aq where q gtp Forj l q the vector 21 is in W Since the columns of B span W the vector 21 is in the column space of B That is a Boy for some vector of of weights Note that c is in RF because B has p columns Let C cl cg Then C is a qu matrix because each of the q columns is in RF By hypothesis q is larger than p so C has more columns than rows By a theorem the columns of C are linearly dependent and there exists a nonzero vector u in R such that Cu 0 P F From part a and the definition of matrix multiplication A a1 aq Bcl Beg BC From part bAu BCu BCu B0 0 Since u is nonzero the columns ofA are linearly dependent IfA contained more vectors than B thenA would be linearly dependent by Exercise 27 because B spans W Repeat the argument with B andA interchanged to conclude that B cannot contain more vectors thanA M Apply the matrix command ref or rref to the matrix v1 v2 x 11 14 19 0 1667 5 8 13 0 C 2667 10 13 18 N 0 0 0 7 10 15 0 0 0 The equation clvl czvz x is consistent so x is in the subspace H The decimal approximations suggest cl 753 and C 83 and it can be checked that these values are precise Thus the Bcoordinate of x is 753 83 30 M Apply the matrix command ref or rref to the matrix v1 v2 v3 x 68 94 003 4 3570 05 97 8 8N002 4 3330000 Chapter 2 Supplementary Exercises 153 The rst three columns of v1 v2 v3 x are pivot columns so v1 v2 and v3 are linearly independent Thus v1 v2 and v3 form a basis B for the subspace H which they span View v1 v2 v3 x as an augmented matrix for 01v1 czv2 C3V3 x The reduced echelon form shows that x is in H and XB 5 2 Notes The Study Guide for Section 29 contains a complete list of the statements in the Invertible Matrix Theorem that have been given so far The format is the same as that used in Section 23 with three columns statements that are logically equivalent for any an matrix and are related to existence concepts those that are equivalent only for any an matrix and those that are equivalent for any nxp matrix and are related to uniqueness concepts Four statements are included that are not in the text s official list of statements to give more symmetry to the three columns The Study Guide section also contains directions for making a review sheet for dimension and rank Chapter 2 SUPPLEMENTARY EXERCISES P P P D FF h39r rare F E P quot5 True IfA andB are an matrices then B7 has as many rows asA has columns so AB7 is defined AlsoATB is de ned becauseAThas m columns andB has m rows False B must have 2 columns A has as many columns as B has rows True The ith row ofA has the form 0 d 0 So the 1th row ofAB is 0 d 0B which is d times the 1th row of B False Take the zero matrix for B Or construct a matrix B such that the equation Bx 0 has nontrivial solutions and construct C and D so that C D and the columns of C 7 D satisfy the equation Bx 0 Then BC7D 0 and BC BD 1 0 0 0 False Counterexample A and C 0 0 0 1 False A BA 7B A2 iAB BA 7B2 This equalsA2 7B2 ifand only ifA commutes with B True An an replacement matrix has n l nonzero entries The ngtltn scale and interchange matrices have n nonzero entries True The transpose of an elementary matrix is an elementary matrix of the same type True An an elementary matrix is obtained by a row operation on 1quot False Elementary matrices are invertible so a product of such matrices is invertible But not every square matrix is invertible True If A is 3 gtlt3 with three pivot positions thenA is row equivalent to 13 False A must be square in order to conclude from the equation AB I thatA is invertible False AB is invertible but AB 1 B 1A71 and this product is not always equal to A lB l True Given AB BA leftmultiply byA 1 to get B A lBA and then rightmultiply by A 1 to obtain BA 1 A IB False The correct equation is rA 1 r lA l because rAr 1A 1 rr 1AA 1 1 I 1 True If the equationAx 0 has a unique solution then there are no free variables in this equation 0 which means thatA must have three pivot positions sinceA is 3X3 By the Invertible Matrix Theorem A is invertible 154 CHAPTER2 MatrixAlgebra 1 1 1 7 5 72 52 200 3 2 000 0 0000 000 3A1001A21001100000 010 010010100 000000 000 A3AAA2100000000 010100 00 Next I AIAA2IAA2 AIAA2IAA2 A A2 7A3 I A3 SinceA3 0 I AAA2I 4 From Exercise 3 the inverse ofI7A is probably A A2 AH To verify this compute I AIAAquot 1IAAquot 1 AIAAquot 1I AAquot 1I Aquot IfAquot 0 then the matrixB 1 A 7A2 Aquot 1 satis es 17AB 1 SinceI7A andB are square they are invertible by the Invertible Matrix Theorem and B is the inverse of 7A 5 A2 2A 7 Multiply byAA3 2A2 7A SubstituteA2 2A 7A3 22A 7 7A 3A 7 2 Multiply by A again A4 A3A 7 2 3A2 7 2A Substitute the identity A2 2A 7 I again FinallyA4 32A 7 7 2A 4A 7 3 Ch 1 0 0 1 2 0 1 LetA 0 1 and B 1 0 BydirectcomputationA IBZIandAB 1 0 7BA gt1 Partial answer in Study Guide Since A713 is the solution ofAX B row reduction of A B to I X will produceX A IB See Exercise 12 in Section 22 138 35138 35138 35 AB2411150 2 5 7757013761 125 34 0 1 36 10 2 57 5 38 35130372910010 1 7013 6 1010 910010 910 001 5 3 001 5 3 001 5 3 10 1 ThusA 1B 910 75 73 8 By definition of matrix multiplication the matrixA satis es Ali iHi ii Chapter 2 Supplementary Exercises 155 2 Rightmultiply both sides by the inverse of 7 The left side becomes A Thus Ar 1 iH i 1 3 5 4 7 3 9 Given AB 2 3 andB2 1notice thatABlBY1 A Since detB 776 1 1 1 a 1 5 4M1 a 3 13 B andAABB 2 7 2 3 2 7 8 27 Note Variants of this question make simple exam questions 10 1 p A p A N 13 SinceA is invertible so is AT by the Invertible lVIatrix Theorem ThenATA is the product of invertible matrices and so is invertible Thus the formula ATA 1AT makes sense By Theorem 6 in Section 22 AT071147 A71AT71ATA711A71 An alternative calculation ATAYIATA AT1f1 ATA I SinceA is invertible this equation shows that its inverse is A TA 1AT Co a For 139 l npq co cm CnilxlquotTl rowiVA rowVc Cnil By a property of matrix multiplication shown after Example 6 in Section 21 and the fact that c was chosen to satisfy Vc y rowiVc rowiVc rowi y y Thus pm y To summarize the entries in Vc are the values of the polynomial px at x1 x 9 Suppose x1 xquot are distinct and suppose Vc 0 for some vector 0 Then the entries in c are the coef cients of a polynomial whose value is zero at the distinct points x1 x However a nonzero polynomial of degree n 7 1 cannot have n zeros so the polynomial must be identically zero That is the entries in 0 must all be zero This shows that the columns of Vare linearly independent 0 Solution in Study Guide When x1 xquot are distinct the columns of Vare linearly independent by b By the Invertible Matrix Theorem Vis invertible and its columns span Rquot So for every y yl yquot in Rquot there is a vector 0 such that Vc y Let p be the polynomial whose coefficients are listed in 0 Then by a p is an interpolating polynomial for x1 yl xn yquot IfA LU then collA LcolllD Since colllD has a zero in every entry except possibly the rst LcollZD is a linear combination of the columns of L in which all weights except possibly the rst are zero So collA is a multiple of collL Similarly colzA LcolzlD which is a linear combination of the columns of L using the rst two entries in colzU as weights because the other entries in colzlD are zero Thus colzA is a linear combination of the first two columns of L a P2 uuTuuT uuTuuT uluT P because u satis es uTu 1 b P7 uuTT uTTuT uuT P c Q2 1 7 2PI 7 2P I 712P 7 2P1 t 213913 17 4P 4P2 1 because ofpart a 156 CHAPTER2 MatriXAlgebra 0 Givenu 0 definePanansinExercise 13 by 1 0 0 0 0 1 0 0 0 0 0 1 0 0 PuuT00 0 10 0 0 QI 2P0 1 01 20 0 00 1 0 1 0 0 l 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 l 0 0 1 1 Ifx 5 thean 0 0 0 5 0 and Qx 0 l 0 5 5 3 0 0 1 3 3 0 0 1 3 3 Equot p A l Leftmultiplication by an elementary matrix produces an elementary row operation B N ElB N EZEIB N E3E2E1B C so B is row equivalent to C Since row operations are reversible C is row equivalent to B Alternatively show C being changed into B by row operations using the inverse of the E SinceA is not invertible there is a nonzero vector v in Rquot such thatAv 0 Place n copies of v into an anmatrixB ThenABAv 8 v Av 8 Av 0 LetA be a 6gtlt4 matrix andB a 4gtlt6 matrix Since B has more columns than rows its six columns are linearly dependent and there is a nonzero x such thath 0 Thus ABx A0 0 This shows that the matrix AB is not invertible by the IMT Basically the same argument was used to solve Exercise 22 in Section 21 Note In the Study Guide It is possible that BA is invertible For example let C be an invertible 4gtlt4 matrix and construct A 18 19 C 0 and B C71 0 Then BA 14 which is invertible By hypothesisA is 5gtlt3 C is 3gtlt5 andAC 13 Suppose x satis es Ax h Then CAx Cb Since CA 1 x must be Cb This shows that Cb is the only solution ofo b 4 2 3 31 26 30 M Let A 3 6 3 Then A2 39 48 39 Instead ofcomputingA3 next speed up the 3 2 4 30 26 31 calculations by computing 2875 2834 2874 2857 2857 2857 A4A2A2 4251 4332 4251 A8A4A4 4285 4286 4285 2874 2834 2875 2857 2857 2857 To four decimal places as k increases 2857 2857 2857 27 27 27 Ak gt 4286 4286 4286 orinrationalformatAk gt 37 37 37 2857 2857 2857 27 27 27 Chapter 2 Supplementary Exercises 157 0 2 3 29 18 18 IfB 1 6 3 then Bl 33 44 33 9 2 4 38 38 49 2119 1998 1998 2024 2022 2022 B4 3663 3764 3663 38 3707 3709 3707 4218 4218 4339 4269 4269 4271 To four decimal places as kincreases 2022 2022 2022 1889 1889 1889 Bk gt 3708 3708 3708 orinrationalformat Bka 3389 3389 3389 4270 4270 4270 3889 3889 3889 20 M The 4gtlt4 matrixA4 is the 4gtlt4 matrix of ones minus the 4gtlt4 identity matrix The MATLAB command is A4 ones4 eye4Fortheinverse use inv A4 70 1 1 1 23 13 13 13 1 0 1 1 1 13 23 13 13 A4 A4 1 1 0 1 13 13 23 13 71 1 1 0 13 13 13 23 70 1 1 1 1 34 14 14 14 14 1 0 1 1 1 14 34 14 14 14 A5 1 1 0 1 1 A571 14 14 34 14 14 1 1 1 0 1 14 14 14 34 14 71 1 1 1 0 14 14 14 14 34 70 1 1 1 1 1 45 15 15 15 15 15 1 0 1 1 1 1 15 45 15 15 15 15 A 1 1 0 1 1 1 A71 15 15 45 15 15 15 6 1 1 1 0 1 1 6 15 15 15 45 15 15 1 1 1 1 0 1 15 15 15 15 45 15 71 1 1 1 1 0 15 15 15 15 15 45 The construction of A6 and the appearance of its inverse suggest that the inverse is related to 15 In fact Agl 16 is 15 times the 6gtlt6 matrix of ones LetJ denotes the an matrix of ones The conjecture is A Jiln and A1AJ 1n n 1 Proof Not required Observe thatJ2 r1 and AnJ J71J J2 7 n 7 1 Now compute Ann 7 1 1J71 n 7 1 1AJ 7A J7 171 1 Since A is squareAn is invertible and its inverse is n 7 1 1J 7 Symmetric Matrices and Quadratic Forms 71 SOLUTIONS Notes Students can profit by reviewing Section 53 focusing on the Diagonalization Theorem before working on this section Theorems 1 and 2 and the calculations in Examples 2 and 3 are important for the sections that follow Note that symmetric matrix means real symmetric matrix because all matrices in the text have real entries as mentioned at the beginning of this chapter The exercises in this section have been constructed so that mastery of the GramSchmidt process is not needed Theorem 2 is easily proved for the 2 x 2 case b IfAa thenkladi a d24b2 c d 2 If I 0 there is nothing to prove Otherwise there are two distinct eigenvalues so A must be diagonalizable d 7 In each case an eigenvector for 7 is b 5 1 Since A 5 7 2 AT the matrix is symmetric 5 2 Since A 3 at AT the matrix is not symmetric 2 3 Since A 4 4 at AT the matrix is not symmetric 8 3 4 Since A 8 O 2 2 AT the matrix is symmetric 2 O 6 2 O 5 Since A O 6 2 at AT the matrix is not symmetric O O 6 6 Since A is not a square matrix A 7t AT and the matrix is not symmetric 379 380 CHAPTER 7 Symmetric Matrices and Quadratic Forms 6 8 7 Let P 8 6 and compute that met Elli 42H le 6 8 Since P is a square matrix P is orthogonal and P71 PT 8 6 15 15 8 Let P 15 15 and compute that T 151515 151o PILL5 15ii15 15ii 1i2 Since P is a square matrix P is orthogonal and P71 PT 15 15 15 1539 5 2 9 Let P 2 5 and compute that T 5 2 5 2 29 0 P P 12 2 5 2 5 0 29 Thus P is not orthogonal 1 2 2 10 Let P 2 l 2 and compute that 2 2 l 1 2 2 1 2 2 900 PTP2 1 2 2 12090 13 2 2 12 2 1009 Thus P is not orthogonal 2 3 2 3 1 3 11 Let P 0 1 5 2J and compute that 53 4N3 2JE 23 0 53 23 23 13 1 0 0 FTP 23 1JE 4JE 0 1J3 2J 0 1 0 13 13 2J 2JE 433 4MB 2J3 0 0 1 12 p n DJ p n A 71 Solutions 381 23 0 JE3 SincePisasquare matriXPis orthogonaland P71PT 23 ls5 4E 13 2J mm 5 5 5 5 5 5 5 5 Let P and compute that 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 1 0 0 0 T 5 5 5 5 5 5 5 5 0 1 0 0 PP 14 5 5 5 5 5 5 5 5 0 0 1 0 5 5 5 5 5 5 5 5 0 0 0 1 5 5 5 1 T 5 5 5 SmceP 1sasquare matr1XP1s orthogonal and P P 5 5 5 5 5 5 3 1 Let A 1 3 Then the characteristic polynomial ofA is 3 7 2 1 7 2 67 8 7 47 2 so 1 the eigenvalues of A are 4 and 2 For 7 4 one computes that a basis for the eigenspace is I which 1 2 1 can be normalized to get 111 For 7 2 one computes that a basis for the eigenspace is I 1 5 1 2 J Let 15 4 0 andD to J Then P orthogonally diagonalizes A and A PDPTI which can be normalized to get 112 1 P u1 zl l 15 15 l 5 Let A 5 I Then the characteristic polynomial ofA is 1 7 2 25 7 2 27 24 7 67 4 1 so the eigenvalues ofA are 6 and 4L For 7 6 one computes that a basis for the eigenspace is 1 15 15 1 I which can be normalized to get 112 which can be normalized to get 111 For 7 41 one computes that a basis for the eigenspace is 1 1N 39 382 p A UI p A Ch p A l LetA 4 1 1 LetA1 3 3 1 CHAPTER7 Symmetric Matrices and Quadratic Forms Let 1 2 l 2 6 0 Pu1 12 J J andD rJi 145 0 4 Then P orthogonally diagonalizes A and A PDP I 16 4 I Then the characteristic polynomial ofA is 16 7 1 7 16 7 2 177 7 177 4 so the eigenvalues ofA are 17 and 0 For 7 17 one computes that a basis for the eigenspace is I 4J 1J which can be normalized to get 111 For 7 0 one computes that a basis for the eigenspace 1 1 17 is which can be normalized to get 112 J Let 4 4J 4 17 1 17 17 0 Pu1 12 J J andD 1J 4J 0 0 Then P orthogonally diagonalizes A and A PDPTI 7 24 Let A 24 7 Then the characteristic polynomial ofA is 7 7 7 7 576 7 2 625 7 257 25 so the eigenvalues ofA are 25 and 725 For 7 25 one computes that a basis for the 3 3 5 eigenspace is 4 which can be normalized to get 111 4 5 For 7 725 one computes that a basis 4 4 5 for the eigenspace is 3 which can be normalized to get 112 35 Let 45 25 0 andD 35 0 25 Then P orthogonally diagonalizes A and A PDPTI 35 45 Pu1 2H 3 1 The eigenvalues of A are 5 2 and 72 For 7 5 one computes that a basis for the 1 1 1 E eigenspace is 1 which can be normalized to get 111 1 J3 For 7 2 one computes that a basis for 1 16 71 Solutions 383 1 1 IE the eigenspace is 2 which can be normalized to get 112 2 J3 For 7 72 one computes that a 1 1 J6 1 1J basis for the eigenspace is 0 which can be normalized to get 113 0 Let 1 1 5 1J 143 1J 5 0 0 Pu1 u2 u3 1J 2JE 0 andD 0 2 0 1J 1JE 1J5 0 0 2 Then P orthogonally diagonalizes A and A PDPTI 2 36 0 18 Let A 36 23 0 The eigenvalues ofA are 25 3 and 750 For 7 25 one computes that a basis 0 0 3 4 4 5 for the eigenspace is 3 which can be normalized to get 111 3 5 For 7 3 one computes that a 0 0 0 0 basis for the eigenspace is 0 which is of length 1 so 112 0 For 7 750 one computes that a 1 1 3 3 5 basis for the eigenspace is 4 which can be normalized to get 113 4 5 Let 0 0 4 5 0 3 5 25 0 0 Pu1 u2 u3 35 0 45 andD 0 3 0 0 1 0 0 0 50 Then P orthogonally diagonalizes A and A PDP I 3 2 4 19 Let A 2 6 2 The eigenvalues of A are 7 and 72 For 7 7 one computes that a basis for the 4 2 3 1 eigenspace is 2 0 This basis may be converted via orthogonal projection to an orthogonal 384 CHAPTER 7 Symmetric Matrices and Quadratic Forms 1 4 l basis for the eigenspace 2 2 Thesevectors canbenormalized to get u1 2MB 5 0 4MB 2 u2 2JB Fork72 one computes thatabasis for the eigenspace is l which can be 5MB 2 23 normalized to get 113 l3 Let 23 1J MIR 23 7 0 0 Pu1 u2 113 2N 2MB 13 andD o 7 o 0 5MB 23 0 0 2 Then P orthogonally diagonalizes A and A PDP I 7 4 4 20 Let A 4 5 0 The eigenvalues of A are 13 7 and 1 For 7 13 one computes that a basis for 4 0 9 2 23 the eigenspace is l which can be normalized to get 111 l 3 For 7 7 one computes that a 2 23 1 l3 basis for the eigenspace is 2 which can be normalized to get 112 2 3 For 7 1 one computes 2 23 2 23 that a basis for the eigenspace is 2 which can be normalized to get 113 2 3 Let 1 l3 23 l3 23 13 0 0 Pu1 112 13 l3 23 23 andD 0 7 0 23 23 l3 0 0 1 Then P orthogonally diagonalizes A and A PDPTI 71 Solutions 385 4 1 3 1 1 4 1 3 21 Let A 3 1 4 1 The e1genvalues ofA are 9 5 and 1 For 7 9 one computes that a bas1s for 1 3 1 4 1 12 1 12 the e1genspace 1s 1 wh1ch can be normallzed to get 111 1 2 For 7 5 one computes that a bas1s 1 12 1 12 1 12 for the e1genspace 1s 1 Wthh can be normallzed to get u2 12 For 7 1 one computes that a 1 12 1 0 0 1 bas1s for the e1genspace 1s 1 0 Thls bas1s 1s an otthogonal bas1s for the e1genspace and these 0 1 45 0 0 1J vectors can be normallzed to get 113 u4 Let 15 0 0 1J2 12 12 1J 0 9 0 0 0 12 12 0 1 2 0 5 0 0 P 1 2 3 4 J andD 12 12 1J2 0 0 0 1 0 12 12 0 1J2 0 0 0 1 Then P otthogonally diagonalizes A and A PDPTI 2 0 0 0 0 0 1 22 Let A 0 0 2 0 The e1genvalues of A are 2 and 0 For 7 2 one computes that a bas1s for the 0 1 0 1 1 0 0 0 1 0 e1genspace 1s 0 0 1 Thls bas1s 1s an otthogonal bas1s for the e1genspace and these vectors 0 1 0 386 CHAPTER 7 Symmetric Matrices and Quadratic Forms 1 0 0 0 1J2 0 can be normallzed to get 111 0 u2 0 and 113 1 For 7 0 one computes that a bas1s for 0 1J5 0 0 0 1 1J the e1genspace 1s 0 wh1ch can be normallzed to get 114 Let 1M 1 0 0 0 2 0 0 0 0120 12 0200 leui 2 3 4 J J andD 0 0 1 0 0 0 2 0 1M 0 1 5 0 0 0 0 Then P orthogonally diagonalizes A and A PDPTI 3 l l 23 Let A l 3 l Since each row ofA sums to 5 1 1 3 1 3 1 1 1 5 1 A1l 3 11551 1 1 1 3 1 5 1 1 lxg and 5 is an eigenvalue ofA The eigenvector 1 may be normalized to get 111 1 J3 One may also 1 1J compute that 1 3 1 1 1 2 1 A l l 3 l l 2 2 l 0 1 1 3 0 0 0 1 so 1 is an eigenvector of A with associated eigenvalue 7 2 For 7 2 one computes that a basis for 0 1 l the eigenspace is l l This basis is an orthogonal basis for the eigenspace and these vectors 0 2 1J 1J can be normalized to get 112 lJ2 and 113 lJg 0 2J3 71 Solutions 387 1J 1J 1J 5 o o Pu1 u2 113 1J lJE 1JE andD 0 2 0 1J 0 2JE 0 0 2 Then P otthogonally diagonalizes A and A PDP I 5 4 2 24 Let A 4 5 2 One may compute that 2 2 2 2 20 2 A 2 20 10 2 1 10 1 2 so v1 2 is an eigenvector ofA with associated eigenvalue XI 10 Likewise one may compute that so 1 is an eigenvector ofA with associated eigenvalue 22 1 For 22 1 one computes that a basis 0 1 1 for the eigenspace is l 0 This basis may be conveIted Via otthogonal projection to an 1 1 otthogonal basis for the eigenspace V2v3 l 1 The eigenvectors v1 v2 and v3 maybe 0 4 23 1J2 lsE normalized to getthevectors u1 23 u2 lJ2 and 113 lxE Let 13 0 MR 23 15 15 10 Pu1 u2 113 23 15 1E andD 0 1 13 0 4JE 0 D OO 0 0 Then P otthogonally diagonalizes A and A PDPTI 388 25 26 DJ 43 CHAPTER 7 Symmetric Matrices and Quadratic Forms True See Theorem 2 and the paragraph preceding the theorem True This is a particular case of the statement in Theorem 1 where u and v are nonzero False There are n real eigenvalues Theorem 3 but they need not be distinct Example 3 apps False See the paragraph following formula 2 in which each u is a unit vector 21 True See Theorem 2 9 True See the displayed equation in the paragraph before Theorem 2 False An orthogonal matrix can be symmetric and hence orthogonally diagonalizable but not every orthogonal matrix is symmetric See the matrix P in Example 2 d True See Theorem 3b O SinceA is symmetric BTABT BTATBTT BT AB and BT AB is symmetric Applying this result with A 1 gives BTB is symmetric Finally BBT T BTTBT BBT so BBT is symmetric LetA be an n X n symmetric matrix Then Ax y Axfy xTATy My x Ay since AT A SinceA is orthogonally diagonalizable A PDP 1 where P is orthogonal andD is diagonal SinceA is invertible A71 PDP 1 1 PD IP I Notice that D 1 is a diagonal matrix so A71 is orthogonally diagonalizable If A and B are orthogonally diagonalizable thenA and B are symmetric by Theorem 2 IfAB BA then ABT BAT ATBT AB So AB is symmetric and hence is orthogonally diagonalizable by Theorem 2 The Diagonalization Theorem of Section 53 says that the columns of P are linearly independent 1 to the of A listed on the diagonal of D So P has exactly k columns of eigenvectors corresponding to 7 These k columns form a basis for the eigenspace If A PRP I then P IAP R Since P is orthogonal R PTAP Hence RT PTAPT PTATPTT PT AP R which shows that R is symmetric Since R is also upper triangular its entries above the diagonal must be zeros to match the zeros below the diagonal Thus R is a diagonal matrix It is previously been found thatA is orthogonally diagonalized by P where 1J 1JE 1J 8 0 0 Pn1 u2 113 1J5 1JE 1J andD0 6 0 0 2JE 1MB 0 0 3 Thus the spectral decomposition of A is T T T T T T A9t1u1u1 9t2u2u2 gtL3U3U3 8u1u1 6u2u2 3u3u3 12 l2 0 16 16 26 13 13 13 8 l2 12 0 6 16 16 26 313 13 13 0 0 0 26 26 46 13 13 13 71 Solutions 389 34 It is previously been found thatA is orthogonally diagonalized by P where 1 5 1E 23 7 0 0 Pu1 u2 u3 0 4JE 13 andD0 7 0 1 5 UM 23 0 0 2 Thus the spectral decomposition of A is T T T T T T Aklu1u1 9L2u2u2 9L3u3u3 7u1u1 7u2u2 2u3u3 U2 0 U2 118 418 ll8 49 29 49 7 0 0 0 7 418 1618 418 2 29 19 29 U2 0 U2 l18 418 118 49 29 49 35 21 Given x in Rquot bx uuT xuuTx uTxu because uTx is a scalar So Bx x uu Since 11 is a unit vector Bx is the orthogonal projection of x onto 11 b Since BT uuT T uTTuT uuT B B is a symmetric matrix Also B2 uuT uuT uuTuuT uuT B because uTu l 0 Since uTu l Bu uuT u uuTu ul u so u is an eigenvector ofB with corresponding eigenvalue 1 36 Given any y in Rquot let y By and z y7 y Suppose that BT B and B2 B Then BTB BB B 21 Since I 9 yfy By y By9 By yTByByYBy yTByyTBTBy 0 z is orthogonal to b Any vector in W Col B has the form Bu for some 11 Noting thatB is symmetric Exercise 28 gives ye Bu Bye 11 Bye93y ll 0 since B2 B So y 79 is in WL and the decomposition y y y 79 expresses y as the sum ofa vector in Wand a vector in Wi By the Orthogonal Decomposition Theorem in Section 63 this decomposition is unique and so y must be projWy 5 2 9 6 2 6 9 37 M Let A 9 6 5 2 The e1genvalues ofA are 18 10 4 and 712 For 7 18 one 6 9 2 5 l l 2 1 l 2 computes that a bas1s for the e1genspace 1s 1 wh1ch can be normallzed to get 111 1 2 For 1 l 2 l l 2 1 l 2 7 10 one computes that a bas1s for the e1genspace 1s 1 wh1ch can be normallzed to get 112 12 l l 2 390 CHAPTER7 DJ MLetA Symmetric Matrices and Quadratic Forms For 7 4 one computes that a basis for the eigenspace is 1 which can be normalized to get 1 12 1 12 I I I 1 I 12 For 7 712 one computes that a bas1s for the e1genspace 1s 1 whlch can be 12 1 12 12 12 12 I Let Pu1 2 3 4 12 12 12 12 12 12 12 12 u3 12 12 12 12 normalized to get 114 O Then P orthogonally diagonalizes A and A PDP I 04 12 06 12 The eigenvalues ofA are 25 30 55 and 75 For 7 25 04 12 one computes that a basis for the eigenspace is For 4 2 2 wh1ch can be normallzed to get 111 1 biplast 1 7t 30 one computes that a basis for the eigenspace is which can be normalized to get 2 2 4 1 112 4 For 7 55 one computes that a basis for the eigenspace is 4 which can be normalized 8 2 4 2 2 4 to get 113 8 For 7 75 one computes that a basis for the eigenspace is 1 which can be 71 Solutions 391 4 8 2 4 4 8 4 4 2 8 normalized to et 11 Let P u u u u and g 4 2 1 2 3 4 4 4 8 2 4 2 8 4 4 25 0 0 0 0 30 0 0 1 D 0 55 0 Then P orthogonally d1agonahzesA and A PDP 0 0 0 75 31 58 08 44 58 56 44 58 M Let A The e1genvalues ofA are 75 0 and 7125 For 7 75 one 08 44 19 08 44 58 08 31 1 3 0 2 computes that a bas1s for the e1genspace 1s 0 2 Th1s bas1s may be converted V1a orthogonal 1 0 1 3 WE 0 4 0 proect10n to the orthogonal bas1s 0 4 These vectors can be normahzed to get 111 0 1 3 WE 3 2 4 50 1 2 J For 7 0 one computes that a basis for the eigenspace is which can be 4 J5 4 3J 2 4 2 2 4 normahzed to get 113 8 For 7 7125 one computes that a bas1s for the e1genspace 1s 4 2 4 8 wh1ch can be normahzed to get 114 2 4 1x2 ME 4 4 75 0 0 0 0 4 50 2 8 0 75 0 0 LetPu1 112 113 14 J and D ThenP 0 4 8 2 0 0 0 0 1J5 3 4 4 0 0 0 125 orthogonally diagonalizes A and A PDPTI 392 CHAPTER 7 Symmetric Matrices and Quadratic Forms 10 2 2 6 9 2 10 2 6 9 40 M Let A 2 2 10 6 9 The eigenvalues ofA are 8 32 728 and 17 For 7 8 one 6 6 6 26 9 9 9 9 9 19 1 1 1 0 computes that a basis for the eigenspace is 0 l This basis may be converted via orthogonal 0 0 0 0 1 1 1 1 projection to the orthogonal basis 0 2 These vectors can be normalized to get 0 0 1JE 1JE 1 1J 1N3 1 111 0 u2 2Jg For 7 32 one computes that a basis for the eigenspace is 1 which 0 0 3 0 0 0 1MB 1MB can be normalized to get 113 1 J5 For 7 728 one computes that a basis for the eigenspace is 3JE 0 1 ME 1 um I which can be normalized to get 114 Um For 7 17 one computes that a basis for the 1 WE 4 mm 1 1 5 1 1J3 eigenspace is l which can be normalized to get 115 1MB 1 1 5 1 1J3 72 Solutions 393 1 5 1JE 15 nm 1MB 1J NE NE WE 1MB LetPu1 112 113 114 us 0 2JE 15 1 1J3 and 0 0 3JE nm 1MB 0 0 0 mm 1MB 8 0 0 0 0 0 8 0 0 0 D 0 0 32 0 0 Then P orthogonally diagonalizes A and A PDPTI 0 0 0 28 0 0 0 0 0 17 72 SOLUTIONS Notes This section can provide a good conclusion to the course because the mathematics here is widely used in applications For instance Exercises 23 and 24 can be used to develop the second derivative test for functions of two variables However if time permits some interesting applications still lie ahead Theorem 4 is used to prove Theorem 6 in Section 73 which in turn is used to develop the singular value decomposition 5 13 x1 1 a XTAXC1 xJLg 1x25c1223x1x2x22 6 b When x xTAx 562 236112 185 1 c When x xTAx 512 231332 16 3 4 3 0 x1 2 21xTAxx1 x2 g 3 2 1 x2 4x122x 9g26x1x22x2Jg 1 In 2 b When x 1 xTAx4222 125262 12 1521 5 1 5 c When x 1J xTAx41J 221J 21J 261J 1J 21J 1J 5 1J S43 a 10 3 The matrix of the quadratic form is 3 3 Uquot 5 3 2 The matrix of the quadratic form is 3 2 0 394 CHAPTER 7 Symmetric Matrices and Quadratic Forms 20 15 2 4 a The matrlx of the quadratlc form 1s 15 2 10 0 1 2 b The matrlx of the quadratlc form 1s 1 2 0 7 8 3 2 5 a The matrix of the quadratic form is 3 7 1 2 1 3 70 2 3 b The matrix of the quadratic form is 2 0 4 73 4 0 5 5 2 3 2 6 a The matrix of the quadratic form is 5 2 1 0 3 2 0 7 7 0 2 0 b The matrix of the quadratic form is 2 0 2 0 2 1 1 5 7 The matrix of the quadratic form is A 5 I The eigenvalues of A are 6 and 41 An eigenvector for 1 1 2 1 7 6 is which may be normalized to 111 J An eigenvector for 7 41 is which may 1 1M 1 1N5 AME and 1N 1 5 l 2 be normalized to 112 Then A PDP 1 where P 111 uz l 2 6 0 D 4 The desired change of variable is x Py and the new quadratic form is xTAxPyTAPyyTPTAPyyTDy6yf 4y 9 4 4 8 The matrix of the quadratic form is A 4 7 0 The eigenvalues ofA are 3 9 and 15 An 4 0 11 2 2 3 eigenvector for 7 3 is 2 which may be normalized to 111 2 3 An eigenvector for 7 9 is 1 1 3 1 1 3 2 2 which may be normalized to u2 23 An eigenvector for 7 15 is 1 which may be 2 3 2 2 72 Solutions 395 23 23 l3 23 normalized to 113 13 Then APDP 1 where P 111 112 13 23 23 13 and 23 13 23 23 0 3 0 D 0 9 0 The desired change of variable is x Py and the new quadratic form is 0 0 15 XTAX PyTAPy yTPTAPy yTDy 3y 9y 15y32 3 2 9 The matrix of the quadratic form is A 2 6 The eigenvalues ofA are 7 and 2 so the quadratic 2JE 2 2 5 An eigenvector for 7 2 is I which may be normalized to 112 Then A PDPTI where 1 5 FYI E 2 5 7 0 and D The desired change of variable is x Py and the 2A5 173 0 2 1 1 5 form is positive de nite An eigenvector for 7 7 is 2 which may be normalized to 111 Piu1 2 new quadratic form is xTAx PyTAPy yTPTAPy My 7y 2y 9 4 10 The matrix of the quadratic form is A 4 3 The eigenvalues ofA are 11 and 1 so the quadratic 2 2 5 form is positive de nite An eigenvector for 7 11 is I which may be normalized to 111 1J 1 1 5 An eigenvector for 7 1 is 2 which may be normalized to 112 Then A PDP 1 where 2 5 2J3 1J3 11 0 P 111 u2 and D The des1red change of var1ab1e 1s x Py and the 473 2 5 0 1 new quadratic form is XTAX PyTAPy yTPTAPy yTDy 11y12 y 2 5 11 The matrlx of the quadrat1c form 1s A 5 2 The e1genvalues of A are 7 and 73 so the quad139at1c 1 1 2 form is indefmite An eigenvector for 7 7 is I which may be normalized to 111 An 1 5 Then A PDP I 1J 145 1 eigenvector for 7 73 is I which may be normalized to u2 396 p A N p A DJ p A A The matrix of the quadratic form is A 3 8 The matrix of the quadratic form is A 3 CHAPTER 7 Symmetric Matrices and Quadratic Forms 1 2 l 2 7 0 where P 111 u2 J l and D The desired change ofvariable is x Py 1A5 1 J5 0 3 and the new quadratic form is xTAxPyTAPyyTPTAPyyTDy7yf 3y 5 2 The matrix of the quadratic form is A 2 2 The eigenvalues of A are 71 and 4 so the quadratic 243 Then A PDP I 1 form is negative de nite An eigenvector for 7 fl is 2 which may be normalized to 111 2J 1 5 l 0 and D 0 6 The desired change of variable is x Py 1M 2 An eigenvector for 7 is I which may be normalized to u2 1N 2J wherePu u 1 2 W3 1 f5 and the new quadratic form is xTAx PyTAPy yTPTAPy yTDy y3 6y l 3 9 The eigenvalues ofA are 10 and 0 so the quadratic 1 form is positive semidefmite An eigenvector for 7 10 is 3 which may be normalized to NM 3 3AM u1 An eigenvector for 7 0 1s which may be normallzed to u2 Then 44 1 W5 A PDP 1 where P 111 15 3 l10 d D 10 0 ll an 2 2 3J 1x10 0 0 variable is x Py and the new quadratic form is xTAx PyTAPy yTPTAPy yTDy 10y12 The desired change of 3 j The eigenvalues of A are 9 and 71 so the quadratic 0 3AM UM 39 Then A PDP 1 where 3 form is indefmite An eigenvector for 7 9 is I which may be normalized to 111 1 3JE l eigenvector for 7 fl is 3 which may be normalized to u2 3 10 1 10 9 0 P 111 u J J and D The desired change of variable is x Py and the WE 3AM 0 1 new quadratic form is xTAxPyTAPyyTPTAPyyTDy9yf y 72 Solutions 397 2 2 2 2 15 M The matrix of the quadratic form is A i 6 g 3 The eigenvalues ofA are 0 76 78 2 0 3 9 and 712 so the quadratic form is negative 39 P quot The 1 quot 39 may be computed 3 0 l 0 20 11 6 21gt1 8 12 12 1 1 1 These eigenvectors may be normalized to form the columns of P and A PDP 1 where 35 0 12 0 0 0 0 0 P 15 2JE 12 0 mm 0 6 0 0 1JE 1JE 12 1J2 0 0 8 0 15 1JE 12 1 0 0 0 12 The desired change of variable is x Py and the new quadratic form is XTAX PyTAPy yTPTAPy yTDy 6y 8y 12y 4 32 0 2 3 2 4 2 0 The eigenvalues ofA are 132 0 2 4 3 2 2 0 3 2 4 and 3 2 so the quadratic form is positive definite The corresponding eigenvectors may be computed 9 M The matrix of the quadratic form is A 4 3 4 3 0 5 0 5 2132 X32 3 4 3 5 0 5 0 Each set of eigenvectors above is already an orthogonal set so they may be normalized to form the columns of P and A PDP 1 where 35 45 35 45 132 0 0 0 55 0 55 0 0 132 0 0 2 45 35 45 35 MD 0 0 32 0 0 55 0 55 0 0 0 32 The desired change of variable is x Py and the new quadratic form is xTAxPyTAPyyTPTAPyyTDyzgyf yz2 y32 y 398 CHAPTER 7 Symmetric Matrices and Quadratic Forms 1 92 0 6 92 1 6 0 The eigenvalues ofA are l72 0 6 1 92 6 0 92 1 and 7132 so the quadratic form is inde nite The corresponding eigenvectors may be computed 17 M The matrix of the quadratic form is A 4 3 4 3 0 5 0 5 X2172 x 132 3 4 3 5 0 5 0 Each set of eigenvectors above is already an orthogonal set so they may be normalized to form the columns of P and A PDPTI where 3N5 AME 3N5 4N5 172 0 0 0 5 0 4N5 0 0 172 0 0 4 3JE 45 3 and 0 0 432 0 0 545 0 5N5 0 0 0 132 The desired change of variable is x Py and the new quadratic form is T T T T T 2 2 2 2 x AxPy APyy P APyy Dy7y17y2 3y3 7y4 ll 6 6 6 6 1 0 0 18 M The matrix of the quadratic form is A 6 0 0 1 The eigenvalues ofA are 17 l 71 6 0 1 0 and 77 so the quadratic form is indefinite The corresponding eigenvectors may be computed 3 0 0 1 1 0 2 1 217 21 X l X 7 1 1 l 1 1 These eigenvectors may be normalized to form the columns of P and A PDPTI where 3 J5 0 0 l 2 17 0 0 0 1JE 0 2JE 12 dB 0 1 0 0 an 1MB 1J 1JE 12 0 0 1 0 1JE 1 5 iJE 12 0 0 0 7 The desired change of variable is x Py and the new quadratic form is XTAX PyTAPy yTPTAPy yTDy 17y12 y y 7y 19 Since 8 is larger than 5 the x term should be as large as possible Since x12 x22 l the largest value that x2 can take is l and x1 0 when x2 1 Thus the largest value the quadratic form can take when xTx1 is 50 81 8 72 Solutions 399 20 Since 5 is larger in absolute value than 73 the x12 term should be as large as possible Since x12 x22 1 22 24 25 the largest value that x1 can take is 1 and x2 0 when x1 1 Thus the largest value the quadratic form can take when xTx 1 is 51 7 30 5 a True See the definition before Example 1 even though a nonsymmetric matrix could be used to compute values of a quadratic form 6quot True See the paragraph following Example 3 O True The columns of P in Theorem 4 are eigenvectors ofA See the Diagonalization Theorem in Section 53 False Qx 0 when x 0 True See Theorem 5a m True See the Numerical Note after Example 6 a True See the paragraph before Example 1 Uquot False The matrix P must be orthogonal and make PT AP diagonal See the paragraph before Example 4 P False There are also degenerate cases a single point two intersecting lines or no points at all See the subsection A Geometric View of Principal Axes d False See the definition before Theorem 5 e True See Theorem 5b If XTAX has only negative values for x 0 then xTAx is negative definite The characteristic polynomial of A may be written in two ways 11 b detA M det b d X X2 adkad b2 and O gt 1gt 2 9 2 0 1 9 2gt 9 19 2 The coef cients in these polynomials may be equated to obtain X1 X2 a d and MM ad b2 det A If detA gt 0 then by Exercise 23 AM gt 0 so that X1 and X2 have the same sign also aaldetAb2 gt0 a If detA gt 0 and a gt 0 then dgt 0 also since adgt 0 By Exercise 23 X1X2 a d gt 0 Since X1 and X2 have the same sign they are both positive So Q is positive de nite by Theorem 5 b If detA gt 0 and a lt 0 then d lt 0 also since adgt 0 By Exercise 23 X1 X2 a d lt 0 Since X1 and X2 have the same sign they are both negative So Q is negative definite by Theorem 5 c If detA lt 0 then by Exercise 23 MM lt 0 Thus X1 and X2 have opposite signs So Q is indefinite by Theorem 5 Exercise 27 in Section 71 showed that BTB is symmetric Also XTBTBX BxT Bx H Bx H 2 0 so the quadratic form is positive semide nite and the matrix ET B is positive semidefinite Suppose that B is square and invertible Then if XTBTBX 0 H Bx 0 and Bx 0 Since B is invertible x 0 Thus if x 0 XTBTBX gt 0 and BTB is positive definite 400 73 CHAPTER 7 Symmetric Matrices and Quadratic Forms Let A PDPT where PT P71 The eigenvalues ofA are all positive denote them awn Let C be the diagonal matrix with JZw Z on its diagonal Then D C2 CTC If B PCPT thenB is positive de nite because its eigenvalues are the positive numbers on the diagonal of C Also BTB PCPT T PCPT PTTCTPT PCPT PCTCPT PDPT A since PT P I Since the eigenvalues ofA and B are all positive the quadratic forms xTAx and xTBx are positive de nite by Theorem 5 Let x 0 Then XTAX gt 0 and XTBX gt 0 so xT A Bx xTAx XTBX gt 0 and the quadratic form XT A Bx is positive definite Note thatA B is also a symmetric matrix Thus by Theorem 5 all the eigenvalues ofA B must be positive The eigenvalues of A are all positive by Theorem 5 Since the eigenvalues of A71 are the reciprocals of the eigenvalues of A see Exercise 25 in Section 51 the eigenvalues of A71 are all positive Note that A 1 is also a symmetric matrix By Theorem 5 the quadratic form XTA IX is positive de nite SOLUTIONS Notes Theorem 6 is the main result needed in the next two sections Theorem 7 is mentioned in Example 2 of Section 74 Theorem 8 is needed at the very end of Section 75 The economic principles in Example 6 may be familiar to students who have had a course in macroeconomics 1 N 5 2 0 The matrix of the quadratic form on the left is A 2 6 2 The equality of the quadratic forms 0 2 7 implies that the eigenvalues of A are 9 6 and 3 An eigenvector may be calculated for each eigenvalue and normalized 13 23 23 29 23 X6 13 X3 23 23 13 13 13 23 23 The desired change ofvariable isxPy where P 23 l3 23 23 23 13 3 l l The matrix of the quadratic form on the left is A l 2 2 The equality of the quadratic forms 1 2 2 implies that the eigenvalues of A are 5 2 and 0 An eigenvector may be calculated for each eigenvalue and normalized 1 5 X51 k2 1J 2JE 0 1JE x0 1J WE 1J2 73 Solutions 401 1J3 2JE 0 The desired change ofvariable is x Py where P lJ3 ls6 12 1A5 1JE 1 5 3 a By Theorem 6 the maximum value of XTAX subject to the constraint XTX 1 is the greatest eigenvalue kl ofA By Exercise 1 219 b By Theorem 6 the maximum value of XTAX subject to the constraint XTX 1 occurs at a unit 1 3 eigenvector 11 corresponding to the greatest eigenvalue XI of A By Exercise 1 u i 2 3 2 3 c By Theorem 7 the maximum value of XTAX subject to the constraints XTX l and xTu 0 is the second greatest eigenvalue 22 of A By Exercise 1 22 6 4 a By Theorem 6 the maximum value of XTAX subject to the constraint XTX 1 is the greatest eigenvalue XI ofA By Exercise 2 XI 5 b By Theorem 6 the maximum value of XTAX subject to the constraint XTX 1 occurs at a unit 1J3 eigenvector 11 corresponding to the greatest eigenvalue XI of A By Exercise 2 u i 1J3 1MB c By Theorem 7 the maximum value of XTAX subject to the constraints XTX l and xTu 0 is the second greatest eigenvalue 22 of A By Exercise 2 22 2 5 5 The matrix of the quadratic form is A 2 5 The eigenvalues ofA are kl 7 and 22 3 a By Theorem 6 the maximum value of XTAX subject to the constraint XTX 1 is the greatest eigenvalue kl ofA which is 7 b By Theorem 6 the maximum value of XTAX subject to the constraint XTX 1 occurs at a unit eigenvector 11 corresponding to the greatest eigenvalue XI of A One may compute that is an 1J 1 Ej c By Theorem 7 the maximum value of XTAX subject to the constraints XTX l and xTu 0 is the second greatest eigenvalue 22 of A which is 3 eigenvector corresponding to XI 7 so u i 7 32 6 The matrix of the quadratic form is A 3 2 3 The eigenvalues of A are XI 15 2 and 22 5 2 a By Theorem 6 the maximum value of XTAX subject to the constraint XTX 1 is the greatest eigenvalue kl ofA which is 152 402 gt1 9 CHAPTER 7 Symmetric Matrices and Quadratic Forms b By Theorem 6 the maximum value of xTAx subject to the constraint xTx 1 occurs at a unit eigenvector 11 corresponding to the greatest eigenvalue XI of A One may compute that I is an 3AM UM 39 c By Theorem 7 the maximum value of XTAX subject to the constraints xTx l and xTu 0 is the second greatest eigenvalue 22 of A which is 52 eigenvector corresponding to XI 7 so u i The eigenvalues of the matrix of the quadratic form are XI 2 22 l and X3 4 By Theorem 6 the maximum value of XTAX subject to the constraint xTx 1 occurs at a unit eigenvector u 1 2 l is an eigenvector 1 corresponding to the greatest eigenvalue XI ofA One may compute that 13 corresponding to 212 so ui 23 23 The eigenvalues of the matrix of the quadratic form are XI 9 and 22 3 By Theorem 6 the maximum value of XTAX subject to the constraint xTx 1 occurs at a unit eigenvector 11 corresponding 1 2 to the greatest eigenvalue XI of A One may compute that 0 and l are linearly independent 1 0 1 eigenvectors corresponding to XI 2 so u can be any unit vector which is a linear combination of 0 1 2 and l Alternatively u can be any unit vector which is quot 39 to the 39 l l quot to 0 1 the eigenvalue 22 3 Since multiples of 2 are eigenvectors corresponding to 22 3 u can be any 1 1 unitvector orthogonal to 2 1 This is equivalent to nding the maximum value of XTAX subject to the constraint xTx 1 By Theorem 6 this value is the greatest eigenvalue XI of the matrix of the quadratic form The matrix of the quadratic 7 1 form 1s A 1 3 and the e1genvalues ofA are XI 5 J3 22 5 Thus the desned constrained maximum value is XI 5 xg H H H 4 Sincex is an 39 73 Solutions 403 This is equivalent to nding the maximum value of xTAx subject to the constraint xTx 1 By Theorem 6 this value is the greatest eigenvalue XI of the matrix of the quadratic form The matrix of the quadratic 3 1 form is A 1 5 and the eigenvalues ofA are kl 1J XZ 1 J Thus the desired constrained maximum value is XI 1 17 A ofA quot to the 39 39 3Ax 3x and XTAX xT3x 3xTx 3 H XHZ 3 since x is aunitvector Letx be a unit eigenvector for the eigenvalue 7 Then XTAX xT Xx MXT x 9 since xTx 1 So 7 must satisfy m S 7 SM Ifm M then let I 1 70m 0M m and x 11 Theorem 6 shows that ugAun m Now suppose thatm ltM and let tbe between m andM Then 0 S rim SMim and 0 S timM7m S 1 Let x t7 mM7m and let x Vl all xEul The vectors ll 0w and xEul are orthogonal because they are eigenvectors for different eigenvectors or one of them is 0 By the Pythagorean Theorem XTXHXHZHMM HZ llEU1HZ110 Hlun HZ 1061HU1HZ105051 since u and 111 are unitvectors and 0 S x S 1 Also since u and 111 are orthogonal XTAX lln Jgu1TA un Eu1 Mun xEu1Tmx un MwEu1 ll almuiunlalMulful1 amaMt Thus the quadratic form XTAX assumes every value between m andM for a suitable unit vector x 0 l 2 3 2 15 l 2 0 15 32 M The matrix of the quadratic form is A The eigenvalues of A are 3 2 15 0 l 2 15 32 1 2 0 xl17 x213 x3 14 and x4 16 a By Theorem 6 the maximum value of XTAX subject to the constraint xTx 1 is the greatest eigenvalue kl ofA which is 17 b By Theorem 6 the maximum value of XTAX subject to the constraint xTx 1 occurs at a unit 1 1 e1genvector u correspondmg to the greatest e1genvalue XI of A One may compute that 1 IS an 1 l 2 l 2 e1genvector corresponding to XI 17 so u i 1 2 l 2 c By Theorem 7 the maximum value of XTAX subject to the constraints xT x1 and xTu 0 is the second greatest eigenvalue M of A which is 13 404 CHAPTER7 Symmetric Matrices and Quadratic Forms 0 32 52 72 32 0 72 5 15 M The matrix of the quadratic form is A The eigenvalues ofA are 52 72 0 32 72 52 32 0 xl152 x 12 x3 52 and X4 92 a By Theorem 6 the maximum value of XTAX subject to the constraint xTx 1 is the greatest eigenvalue X1 ofA which is 152 b By Theorem 6 the maximum value of XTAX subject to the constraint xTx 1 occurs at a unit l e1genvector u correspondmg to the greatest e1genvalue ll of A One may compute that eigenvector corresponding to X1 15 2 so u is an 1 1 12 12 12 12 c By Theorem 7 the maximum value of XTAX subject to the constraints xTx l and xTu 0 is the second greatest eigenvalue X2 of A which is 712 4 3 16 M The matrix of the quadratic form is A 5 5 X2 3 X3 1 and X4 9 a By Theorem 6 the maximum value of XTAX eigenvalue X1 ofA which is 9 b By Theorem 6 the maximum value of XTAX eigenvector 11 corresponding to the greatest eigenvalue ll of A One may compute that eigenvector corresponding to X1 9 so u i 3 5 5 0 3 3 The e1genvalues of A are X1 9 3 0 l 3 l 0 subject to the constraint KT 1 is the greatest subject to the constraint KT 1 occurs at a unit 2 is an 2JE 0 1J3 39 1 c By Theorem 7 the maximum value of XTAX subject to the constraints XTX l and xTu 0 is the second greatest eigenvalue X2 of A which is 3 74 Solutions 405 6 2 2 2 0 17 M The matrix of the quadratic form is A 2 0 13 The eigenvalues of A are X1 4 2 0 3 13 X2 10 X3 12 and X4 16 a By Theorem 6 the maximum value of XTAX subject to the constraint xTx 1 is the greatest eigenvalue X1 ofA which is 41 subject to the constraint xT x 1 occurs at a unit b By Theorem 6 the maximum value of XTAX 3 eigenvector 11 corresponding to the greatest eigenvalue ll of A One may compute that is an 3JE 1JE eigenvector corresponding to 9 4 so u i 1 1MB 1JE c By Theorem 7 the maximum value of XTAX subject to the constraints xTx 1 and xTu 0 is the second greatest eigenvalue X2 of A which is 710 74 SOLUTIONS Notes The section presents a modern topic of great importance in applications particularly in computer calculations An understanding of the singular value decomposition is essential for advanced work in science and engineering that requires matrix computations Moreover the singular value decomposition explains much about the structure of matrix transformations The SVD does for an arbitrary matrix almost what an orthogonal decomposition does for a symmetric matrix 1 0 1 0 1 Let A Then ATA and the eigenvalues of ATA are seen to be in decreasing 3 0 9 order X1 9 and X2 1 Thus the singular values ofA are 0391 J9 3 and 0392 J1 1 5 0 T 25 2 LetA ThenAA 0 0 0 order X1 25 and X2 0 Thus the singular values ofA are 01JE 5 and 0392 J6 0 J6 0 and the eigenvalues of AT A are seen to be in decreasing 1 T 6 ThenA A J3 J3 X2 13gt 36 X 9gt 4 and the eigenvalues of ATA are in decreasing order X1 9 and X2 4 J3 T 3 Let A and the character1st1c polynomlal of A A 1s Thus the singular values ofA are 0391 I9 3 and 0392 J4 2 406 CHAPTER 7 Symmetric Matrices and Quadratic Forms 4 Let A JE 2J 7 X2 10gt 9 X 9gt l and the eigenvalues of ATA are in decreasing order X1 9 and X2 1 Thus the singular values ofA are 0391 J9 3 and 0392 Jl l 5 2 32J Then AT A J and the characteristic polynomial of AT A is UI 3 0 9 0 Let A 0 0 Then ATA 0 0 and the eigenvalues of ATA are seen to be in decreasing order X1 9 and X2 0 Associated unit eigenvectors may be computed 1 0 9 9 9 0 0 l l 0 Thus one cholce for Vis V 0 1 The s1ngular values ofA are 0391 J9 3 and 0392 J6 0 Thus 3 0 the matrlx 2 1s 2 0 0 Next compute 1 11 LA 0391 0 Becausesz 0 the only column found for U so far is 111 Find the other column of U is found by extending 111 to an orthonormal basis for R2 An easy choice is uz l 0 Let U Thus 0 l tm m 2 0 4 0 Let A 0 I Then ATA 0 I and the eigenvalues of ATA are seen to be in decreasing Ch order X1 4 and X2 1 Associated unit eigenvectors may be computed Mimi 0 1 Thus one choice for Vis V 0 The singular values ofA are 0391 J4 2 and 0392 xT 1 Thus 2 0 the matrix 2 is E 0 1 Next compute 1 1 1 0 u1 Av1 u2 Av2 0391 0 0392 1 l 0 Since u1u2 is abasis for R2 let U 0 1 Thus tm j m 74 Solutions 407 l 2 l 8 2 Let A 2 2 Then AT A 2 5 and the characteristic polynomial of AT A is W 13gt 36 X 9X 4 and the eigenvalues of ATA are in decreasing order XI 9 and M 4 Associated unit eigenvectors may be computed 2J 1J XJinxElJ zli 2J3l 243 1J Thus one choice for Vis V The singular values ofA are 039 J6 3 and ins3 2A5 1 3 0 0392 J4 2 Thus the matr1x 2 1s 2 0 2 Next compute ulAvlJguIAv 2J 101 1 2 2 02 2 1M 1 5 2 5 Since u1u2 is abasis for R2 let U J J Thus 2A5 1 3 0 26 1A5 0 2 1J 2J3 3 4 6 Let A 0 2 Then AT A 6 13 and the characteristic polynomial of AT A is A UEVT 1N3 2 E 2N 1J3 on W 17gt16 X 16gt 1 and the eigenvalues of ATA are in decreasing order XI 16 and M 1 Associated unit eigenvectors may be computed 1J3 2J MilAlli M 16 2J Thus one choice for Vis V The singular values ofA are 0391 JR 4 and 2 13 1M 4 0 0392 xl 1 Thus the matrix 2 is Z 0 1 Next compute 2 5 l 5 I 01 1J3 02 2J3 2 5 1 5 Since u1u2 isabasis for R2 let U J J Thus 1J3 2MB 25 15401525 A21 sz J J J J 1J3 2 5 0 1 2J 143 408 CHAPTER 7 Symmetric Matrices and Quadratic Forms 7 l 74 3 9 Let A 0 0 Then ATA 32 26 and the characteristic polynomial of ATA is 5 5 X2 100gt4 900 X 90X 10 and the eigenvalues of ATA are in decreasing order X1 90 and X2 10 Associated unit eigenvectors may be computed 2 5 l 5 X90 J X10 J 1J5 2 5 2 5 1 5 Thus one choice for Vis V J J The singular values ofA are 0391 m 3m and 1 5 2MB 3J5 0 0392 m Thus the matrix 2 is Z 0 x Next compute 0 0 1J2 1J u1iAv1 0 u2 Av2 0 0391 1A5 0392 1 5 Since u1u2 is not a basis for R3 we need a unit vector 113 that is orthogonal to both 111 and Hz The vector 113 must satisfy the set of equations if X 0 and u x 0 These are equivalent to the linear equations 0 0 xl0x2x3 0sox landu3 l x10xzx3 0 0 145 1J 0 Therefore let U 0 0 l Thus 145 145 0 1JE 1J 0 3J5 0 2 5 l 5 AU2VT 0 0 1 0 JE f l 5 2 5 1H 1A5 0 0 0 4 2 T 20 10 T 10 Let A 2 l Then A A 10 5 and the characteristlc polynomial ofA A 1s 0 0 X2 25 M 25 and the eigenvalues of ATA are in decreasing order X1 25 and X2 0 Associated unit eigenvectors may be computed 2N3 1J i251 i02 74 Solutions 409 2 5 1 5 Thus one choice for Vis V J J The singular values ofA are 0391 IE 5 and 1J 2J3 5 0 0392 J6 0 Thus the matrix 2 is Z 0 0 Next compute 0 0 2M5 u1 iAv1 lsg 0391 0 Becausesz 0 the only column found for U so far is 111 Find the other columns of U found by extending 111 to an orthonormal basis for R3 In this case we need two orthogonal unit vectors uz and u3 that are orthogonal to 111 Each vector must satisfy the equation If x 0 which is equivalent to the equation 2x1 x2 0 An orthonormal basis for the solution set of this equation is 1MB 0 Hz 2 u3 0 0 1 Therefore letU lJg 2J 0 2MB 143 0 Thus 0 0 1 26 1J3 0 5 0 I J 2 5 1 5 AU2VT 1A5 2J 0 0 0 J 0 0 1 0 0 16 2N3 3 1 81 11 LetA 6 2 Then ATA 27 9 and the characteristic polynomial of AT A is 6 2 X2 90gt MX 90 and the eigenvalues of ATA are in decreasing order X1 90 and X2 0 Associated unit eigenvectors may be computed 3 10 1 10 9t90 J X0 J 1E 3MB 3 10 1 10 Thus one choice for Vis V J J The singular values ofA are 0391 3m and 1J10 345 3J5 0 0 0 Next compute 0 0 0392 J6 0 Thus the matrix 2 is Z 1 13 111 Av1 23 71 23 410 CHAPTER 7 Symmetric Matrices and Quadratic Forms Becausesz 0 the only column found for U so far is 111 The other columns of U can be found by extending 111 to an orthonormal basis for R3 In this case we need two orthogonal unit vectors uz and u3 that are orthogonal to 111 Each vector must satisfy the equation if x 0 which is equivalent to the equation x1 2x2 2x3 0 An orthonormal basis for the solution set of this equation is 23 23 2 13 u3 23 23 l3 l3 23 23 Therefore let U 23 l3 23 Thus 23 23 13 13 23 23 310 0 T 310 110 AU2V 23 13 23 23 23 13 0 0 V10 3 0 1 1 2 0 12 Let A 0 l Then ATA 0 3 and the eigenvalues of ATA are seen to be in decreasing order 1 1 XI 3 and 22 2 Associated unit eigenvectors may be computed Hmmm 0 1 Thus one choice for Vis V 1 0 The singular values ofA are 0391 J3 and 0392 Thus the 1 CEO matrix 2 is 2 Next compute 0 0 WE WE u1iAv1 lJ3 u2iAv2 0 01 02 16 1J Since u1u2 is not a basis for R3 we need a unit vector 113 that is orthogonal to both 111 and Hz The vector 113 must satisfy the set of equations if X 0 and u x 0 These are equivalent to the linear equations 0 1 lJg sox 2 andu3 2g 0 1 1J3 951x29 x10x2 g 74 Solutions 411 1J3 1J2 1JE Therefore let U 1J3 0 2g Thus 1J3 1J2 1JE 1J3 1N5 1JE J 0 AU2VT1J3 0 2JE 0 J2 i 3 1J3 1J2 MR 0 3 3 2 2 T 13 LetA ThenA 2 2 3 2 2 17 8 3 ATTAT AAT 8 17 and the eigenvalues of ATTAT 2 are seen to be in decreasing order X1 25 and X2 9 Associated unit eigenvectors may be computed 1 2 1 2 9 25 J 9 9 J 1J2 1 1 2 1 2 Thus one choice for Vis V J J The singular values of AT are 039 JE5 and 1J2 ME 1 5 0 0392x 3Thusthematrixis2 0 3 Next compute 0 0 1N2 1 u1iATv1 lJ2 uz iATv2 lxE 0391 0 0392 4JE Since u1u2 is not a basis for R3 we need a unit vector 113 that is orthogonal to both 111 and Hz The vector 113 must satisfy the set of equations If x 0 and u x 0 These are equivalent to the linear equations 0 0 2 23 x1 x2 4x3 0sox 2 andu3 23 xl x2 x3 1 13 1J2 AME 23 Therefore let U ls2 lJl S 23 Thus 0 4J 13 1J2 1E 23 5 ATU2VT1J2 IJE 23 0 0 mm 13 0 0 0 3 1J2 1 5 1J2 1 5 412 p A A EJ39 9 p A l p A o0 gt0 From Exercise 7 A UEVT with V CHAPTER 7 Symmetric Matrices and Quadratic Forms An SVD forA is computed by taking transposes 1 5 15 0 1 2 1 2 5 0 0 r IM WE 4JE 23 13 1 15 15 0 3 0 23 2J3 1J 143 2N3 associated with the greatest eigenvalue XI of AT A so the first column of Vis a unit vector at which Since the rst column of Vis unit eigenvector H Ax H is maximized a SinceA has 2 nonzero singular values rankA 2 40 78 58 b By Example 6 u1u2 37 33 is a basis for ColA and v3 58 is a basis 84 52 58 for NulA a SinceA has 2 nonzero singular values rankA 2 65 34 86 11 08 42 b By Example 6 u1u2 3l 68 1s a bas1s for ColA and V3V4 16 84 1s 41 73 I I 73 08 a basis for Nul A Let A UEVT U E V71 SinceA is square and invertible rankA n and all of the entries on the diagonal of must be nonzero So A71UZ V71 1 V271 U71 V271 UT First note that the determinant of an orthogonal matrix is i1 because 1 detI detU T U detUT detU detU2 Suppose thatA is square and A UZVT Then 2 is square and detA detUdetZdet VT idetZ io l 0 Since U and Vare orthogonal matrices ATA U VTT U2 VT VZTUTU VT VgtTgtVT VZTZV 1 If 03039 are the diagonal entries in 2 then if is a diagonal matrix with diagonal entries 0120392 and possibly some zeros Thus Vdiagonalizes AT A and the columns of Vare eigenvectors of AT A by the Diagonalization Theorem in Section 53 Likewise AAT UgtVTUgtVTT UgtVTVgtT UT UgtgtTUT UZETU 1 so U diagonalizes AAT and the columns of U must be eigenvectors of AAT Moreover the Diagonalization Theorem states that 03912 2 o are the nonzero eigenvalues of ATA Hence 03039 are the nonzero singular values of A If A is positive de nite then A PDPT where P is an orthogonal matrix and D is a diagonal matrix The diagonal entries of D are positive because they are the eigenvalues of a positive de nite matrix Since P is an orthogonal matrix PPT I and the square matrix PT is invertible Moreover From the proof of Theorem 10 U2 o lu1 74 Solutions 413 PT 71 P71T1 P PT T so PT is an orthogonal matrix Thus the factorization A PDPT has the properties that make it a singular value decomposition Let A U2 VT The matrix PU is orthogonal because P and U are both orthogonal See Exercise 29 in Section 62 So the equation PA PU 2 VT has the form required for a singular value decomposition By Exercise 19 the diagonal entries in 2 are the singular values of PA The right singular vector v1 is an eigenvector for the largest eigenvector XI of AT A By Theorem 7 in Section 73 the second largest eigenvalue 22 is the maximum of xT ATAX over all unit vectors orthogonal to v1 Since xT ATAX Ax HZ the square root of 22 which is the second largest singular value of A is the maximum of H Ax H over all unit vectors orthogonal to v1 an 0 0 The columnrow expansion of the product U2VT shows that T v1 A UgtVT U2 o lu1v1T auvT vVL where r is the rank of A 0 for i 39 From Exercise 23 AT o lvlulT o vu Then since ulTu j l forij T T T T T A u 0391v1u1 o vuu 039ijuj u 039Jvuju039v Consider the SVD for the standard matrixA of T say A U2 VT Let B v1vn and C u1 um be bases for Rquot and Rm constructed respectively from the columns of Vand U Since the columns of Vare orthogonal VTV e J where e J is the jth column of the n X n identity matrix To find the matrix of T relative to B and C compute Tv Av U2VTv U2e quj 0Ue 0w so Tv C O39JeJ Formula 4 in the discussion at the beginning of Section 54 shows that the diagonal matrix 2 is the matrix of T relative to B and C 18 13 4 4 528 392 224 176 2 19 4 12 T 392 1092 176 5 6 M Let A Then A A and the e1genvalues 14 11 12 8 224 176 192 128 2 21 4 8 176 536 128 288 of AT A are found to be in decreasing order 21 1600 22 400 unit eigenvectors may be computed X3 100 and X4 0 Associated 4 8 4 2 8 4 2 4 X1 2 XZ 4 3 8 A 4 4 2 4 8 414 CHAPTER 7 Symmetric Matrices and Quadratic Forms 4 8 4 2 8 4 2 4 Thus one cholce for V1s V 2 4 8 4 The Singular values of A are 0391 40 0391 20 4 2 4 8 40 0 0 0 0 20 0 0 0393 10 and 0394 0 Thus the matrle 1s 2 0 10 0 Next compute 0 0 0 0 5 5 5 1 5 u1 Av1 u2 Av2 0391 5 0392 5 5 5 5 l 5 113 Av3 5 5 BecauseAv4 0 only three columns of U have been found so far The last column of U can be found by extending u1 uz 13 to an orthonormal basis for R4 The vector u must satisfy the set of equations if x 0 11 X 0 and nix 0 These are equivalent to the linear equations 1 5 x1x2x3x40 1 5 x1x2 x3x40sox 1andu4 395 x x 0 I x1 2 4 1 5 5 5 5 5 5 5 5 5 Therefore let U Thus 5 5 5 5 5 5 5 5 5 5 5 5 40 0 0 0 4 8 2 4 AzUEVT 5 5 5 5 0 20 0 0 8 4 4 2 5 5 5 5 0 0 10 0 4 2 8 4 5 5 5 5 0 0 0 0 2 4 4 8 41 32 3 8 l4 8 32 118 3 92 74 27 M Let A 0 Then ATA 38 3 121 10 52 and the 14 92 10 81 72 8 74 52 72 100 eigenvalues of AT A are found to be in decreasing order kl 27087 M 14785 X3 2373 X4 1855 and XS 0 Associated unit eigenvectors may be computed 1 244 8 74 Solutions 415 10 39 74 41 36 61 29 27 50 48 XI 21 XZ 84 X3 07 914 45 Xs 19 52 14 38 23 72 55 19 49 58 29 10 39 74 41 36 61 29 27 50 48 Thus one choice for Vis V 21 84 07 45 19 The nonzero singular values ofA are 52 14 38 23 72 55 19 49 58 29 011646 0391 12 16 0393 487 and 0394 431 Thus the matrix 2 is 1646 0 0 0 0 0 1216 0 0 0 Z Next compute 0 0 487 0 0 0 0 0 431 0 57 65 1 63 1 24 u1 Av1 u2 Av2 0391 07 0392 63 51 34 42 27 68 1 29 u3 Av3 u4 AV4 0393 53 0394 56 29 73 57 65 42 27 63 24 68 29 Since u1u2u3u4 is abasis for R4 let U Thus 07 63 53 56 51 34 29 73 AU2VT 57 65 42 27 1646 0 0 0 63 24 68 29 0 1216 0 0 07 63 53 56 0 0 487 0 51 34 29 73 0 0 0 431 39 29 84 14 19 49 H o o o o l 4 N l o 1 Ln oo 36 48 19 72 29 416 28 75 CHAPTER 7 Symmetric Matrices and Quadratic Forms 4 0 7 7 102 43 27 52 6 1 11 9 43 30 33 M Let A Then AT A and the eigenvalues of 7 5 10 19 27 33 279 335 1 2 3 1 52 88 335 492 ATA are found to be in decreasing order X1 7499785 X2 1462009 X3 68206 and X4 13371gtlt10 6 The singular values ofA are thus 0391 273857 0392 120914 0393 261163 and 0394 00115635 The condition number 03910394 23683 5 3 1 7 9 255 168 90 160 47 6 4 2 8 8 168 111 60 104 30 M Let A 7 5 3 10 9 Then ATA 90 60 34 39 8 and the eigenvalues 9 6 4 9 5 160 104 39 415 178 8 5 2 11 4 47 30 8 178 267 of ATA are found to be in decreasing order X1 672589 X2 280745 X3 127503 X4 1163 and X5 1428gtlt10 7 The singular values ofA are thus 0391 259343 0392 167554 0393 112917 0394 107853 and 0395 000377928 The condition number 03910395 68622 SOLUTIONS Notes The application presented here has turned out to be of interest to a wide variety of students including engineers I cover this in Course Syllabus 3 described above butI only have time to mention the idea brie y to my other classes 1 The matrix of observations is X 2 2 20 13 5 19 226 3 and the sample mean is 12 6 9 15 72 12 M The meandeviation form B is obtained by subtractingM from each column of X so 6 60 10 7 10 6 9 10 8 B The sample covarlance matrix 1s 2 4 1 5 3 5 430 135 86 27 S LBBT l 6 1 5 135 80 27 16 1 5 2 6 7 3 3 11 6 8 15 11 The meandeviation form B is obtained by subtractingM from each column of X so 3 1 2 2 3 1 B 62 24 4 The matrix of observations is X and the sample mean is M LJ 3 16 2 1 T 1 28 40 56 8 S BB 6 1 5 40 90 8 18 J The sample covariance matrix is 3 A 9 F 75 Solutions 417 The principal components of the data are the unit eigenvectors of the sample covariance matrix S One 86 27 are 21952041 and 27 16 computes that in descending order the eigenvalues of S 293348 22 679593 One further computes that corresponding eigenvectors are v1 1 and 340892 v 2 These vectors may be normalized to nd the principal components which are 322659 946515 946515 u1 for 21952041 and 112 322659 for x 679593 The principal components of the data are the unit eigenvectors of the sample covariance matrix S One 6 8 are 21219213 and 8 18 computes that in descending order the eigenvalues of S 490158 22 167874 One further computes that corresponding eigenvectors are v1 1 and 204016 2 1 These vectors may be normallzed to nd the pr1nc1pal components wh1ch are 897934 44013 44013 u1 for XI 219213 and 112 897934 J for 22 167874 16412 3273 8104 3273 53944 24913 is XI 677497 and the rst principal 8104 24913 18911 M The largest eigenvalue of S 1295 54 component of the data is the unit eigenvector corresponding to kl which is 111 874423 The fraction 467547 of the total variance that is contained in this component is kl trS 677497 164 12 53944 18911 758956 so 758956 of the variance of the data is contained in the first principal component 2964 1838 500 M The largest eigenvalue of S 1838 2082 1406 is XI 516957 and the first principal 500 1406 2921 615525 component of the data is the unit eigenvector corresponding to kl which is 111 599424 Thus one 511683 choice for the new variable is y1 615525x1 599424x2 511683x3 The fraction of the total variance that is contained in this component is kl trS 5169572964 2082 2921 648872 so 648872 of the variance of the data is explained by yl 418 CHAPTER 7 Symmetric Matrices and Quadratic Forms 946515 322659 variable is y1 9465l5x1 322659x2 The fraction of the total variance that is contained in this component is X1 trS 95204186 16 933374 so 933374 of the variance of the data is explained by yl 7 Since the unit eigenvector corresponding to X1 952041 is 111 one choice for the new 44013 897934 variable is y1 44013x1 897934x2 The fraction of the total variance that is contained in this component is XI trS 2l92l356 18 928869 so 928869 ofthe variance of the data is explained by yl 8 Since the unit eigenvector corresponding to X1 219213 is 111 one choice for the new gt0 5 2 0 The largest eigenvalue of S 2 6 2 is X1 9 and the rst principal component of the data is the 0 2 7 1 3 unit eigenvector corresponding to X1 which is 111 2 3 Thus one choice for y is 2 3 y l 3x1 23x2 23g and the variance ofy is X1 9 5 4 2 10 M The largest eigenvalue of S 4 ll 4 is X1 15 and the first principal component of the data 2 4 5 143 is the unit eigenvector corresponding to X1 which is 111 2 J3 Thus one choice for y is 143 y l JEx1 2xgx2 lJEg and the variance ofy is X1l5 11 a Ifw is the vector in RNwith a l in each position then X1 XNW X1XN 0 since the X k are in meandeviation form Then Y1 YNw PTX1 PTXNJW 13T x1 XNw PT00 Thus Y1 YN 0 and the Yk are in meandeviation form b By part a the covariance matrix SY of Y1YN is 1 Sf m YNY1 YNT 1 EPTX1 XNPTX1 xNT PT X1 xNx1 XNTjPPTSP since the X k are in meandeviation form Chapter 7 Supplementary Exercises 419 12 By Exercise 11 the change of variables X PY changes the covariance matrix S of X into the covariance matrix PT SP of Y The total variance of the data as described by Y is trPT SP However since PT SP is similar to S they have the same trace by Exercise 25 in Section 54 Thus the total variance of the data is unchanged by the change of variables X FY 13 Let M be the sample mean for the data and let 9 Xr M Let B X1 ZN be the matrix of observations in meandeviation form By the rowcolumn expansion of BET the sample covariance matrix is S LEET N l l EB XN A T XN l N A A T 1 N T IZXkXJr EZO9 MXX M k1 k1 Chapter 7 SUPPLEMENTARY EXERCISES P 39 5 01 Pquot a True This is just part of Theorem 2 in Section 71 The proof appears just before the statement of the theorem 1 0 True This is proved in the rst part of the proof of Theorem 6 in Section 73 It is also a consequence of Theorem 7 in Section 62 0 False A counterexample is A 1 False The principal axes of xTAx are the columns of any orthogonal matrixP that diagonalizes A Note WhenA has an eigenvalue whose eigenspace has dimension greater than 1 the principal axes are not uniquely determined 1 1 False A counterexample is P 1 The columns here are orthogonal but not orthonormal 1 False See Example 6 in Section 72 2 0 1 False A counterexample is A 0 3 and x Then xTAx 2 gt 0 but xTAx is an indefinite quadratic form True This is basically the Principal Axes Theorem from Section 72 Any quadratic form can be written as xTAx for some symmetric matrixA False See Example 3 in Section 73 False The maximum value must be computed over the set of unit vectors Without a restriction on the norm of x the values of xTAx can be made as large as desired 420 CHAPTER7 43 UI Ch Symmetric Matrices and Quadratic Forms k False Any orthogonal change of variable x Py changes a positive de nite quadratic form into another positive de nite quadratic form Proof By Theorem 5 of Section 72 the classi cation of a quadratic form is determined by the eigenvalues of the matrix of the form Given a form XTAX the matrix of the new quadratic form is P IAP which is similar to A and thus has the same eigenvalues as A False The term de nite eigenvalue is unde ned and therefore meaningless True Ifx Py then xTAx PyT APy yTPTAPy yTP lAPy 1 71 n False A counterexample is U 1 I The columns of U must be orthonormal to make UUTX the orthogonal projection of x onto Col U 0 True This follows from the discussion in Example 2 of Section 74 which refers to a proof given in Example 1 p True Theorem 10 in Section 74 writes the decomposition in the form UZVT where U and Vare orthogonal matrices In this case VT is also an orthogonal matrix Proof Since Vis orthogonal V is invertible and V71 VT Then VT 71 V71T VT T and since Vis square and invertible VT is an orthogonal matrix 2 0 q False A counterexample is A 0 I The singular values of A are 2 and l but the singular values of ATA are 4 and l a Each term in the expansion ofA is symmetric by Exercise 35 in Section 7 l The fact that B CT BT CT implies that any sum of symmetric matrices is symmetric soA is symmetric b Since ulTu1l and u u10 forj l Au1 9L1u1u1Tu1 knunu u1 X1u1u1Tu1 knunuu1 klul Since 111 0 X1 is an eigenvalue of A A similar argument shows that X is an eigenvalue of A for j2 n If rankA r then dimNulA n 7 r by the Rank Theorem So 0 is an eigenvalue of A with multiplicity n 7 r and of the n terms in the spectral decomposition of A exactly n 7 r are zero The remaining r terms which correspond to nonzero eigenvalues are all rank 1 matrices as mentioned in the discussion of the spectral decomposition a By Theorem 3 in Section 61 ColAi NulAT Nul A since AT A b Let y be in Rquot By the Orthogonal Decomposition Theorem in Section 63 y y z where y is in ColA and z is in ColAL By part a z is in NulA If Av 7w for some nonzero 7 then v ATIAV Ale which shows that v is a linear combination of the columns of A BecauseA is symmetric there is an orthonormal eigenvector basis u1 u for Rquot Let r r39ankA Ifr 0 thenA O and the decomposition of Exercise 4b is y 0 y for each y in Rquot ifr n then the decomposition is y y 0 for each y in Rquot Assume that 0 lt r lt n Then dimNulA n 7 r by the Rank Theorem and so 0 is an eigenvalue ofA with multiplicity n 7 r Hence there are r nonzero eigenvalues counted according to their multiplicities gt1 on H H 12 Chapter 7 Supplementary Exercises 421 Renumber the eigenvector basis if necessary so that u1 u are the eigenvectors corresponding to the nonzero eigenvalues By Exercise 5 u1u are in ColA Also u1un are in NulA because these to the 39 39 A I39 vectors are 39 that 0 For y in Rquot there are scalars 01cn such yclu1cu c1u1cnun A A 5 z This provides the decomposition in Exercise 4b If A RT R and R is invertible thenA is positive de nite by Exercise 25 in Section 72 Conversely suppose thatA is positive de nite Then by Exercise 26 in Section 72 A BT B for some positive de nite matrix B Since the eigenvalues of B are positive 0 is not an eigenvalue of B and B is invertible Thus the columns of B are linearly independent By Theorem 12 in Section 64 B QR for some n X n matrix Q with orthonormal columns and some upper triangular matrix R with positive entries on its diagonal Since Q is a square matrix QT Q I and A BTB QRT QR RTQTQR RTR and R has the required properties Suppose thatA is positive de nite and consider a Cholesky factorization of A RT R with R upper triangular and having positive entries on its diagonal Let D be the diagonal matrix whose diagonal entries are the entries on the diagonal of R Since rightmultiplication by a diagonal matrix scales the columns of the matrix on its left the matrix L RT D71 is lower triangular with 1 s on its diagonal If U DR then A RTD IDR LU IfA is an m X n matrix and x is in Rquot then xTATAx AxT Ax H Ax H2 2 0 Thus ATA is positive sernide nite By Exercise 22 in Section 65 rank ATA rankA If rank G r then dimNul G n 7 r by the Rank Theorem Hence 0 is an eigenvalue of G with multiplicity n 7 r and the spectral decomposition of G is G klululf kruruf Also lbw are positive because G is positive semidefinite Thus G un1Zulfmfuquf By the columnrow expansion of a matrix product G BBT where B is the n X r matrix BUZu1 FinallyGATA forABT Let A UEVT be a singular value decomposition ofA Since U is orthogonal UTU 1 and A UZUTUVT PQ where PUEUT UZU 1 and QUVT Since is symmetricP is symmetric and P has nonnegative eigenvalues because it is similar to E which is diagonal with nonnegative diagonal entries Thus P is positive semide nite The matrix Q is orthogonal since it is the product of orthogonal matrices 21 Because the columns of V are orthonormal My UDVTVD 1UT y UDD 1UT y w y 422 CHAPTER7 Uquot 6quot O Symmetric Matrices and Quadratic Forms Since U U T y is the orthogonal projection of y onto Col U by Theorem 10 in Section 63 and since Col U Col A by 5 in Example 6 of Section 74 AAy is the orthogonal projection of y onto ColA Because the columns of U are orthonormal AAx VD 1Uf UDVTx VD 1DVTx 14fo Since VVT x is the orthogonal projection of x onto Col V by Theorem 10 in Section 63 and since Col V Row A by 8 in Example 6 of Section 74 AAx is the orthogonal projection of x onto Row A Using the reduced singular value decomposition the de nition of A and the associativity of matrix multiplication gives MA UDVTVD 1UT UDVT UDD 1UT UD V3 UDD 1DVT UD VT A AMA KD 1U3UDKTKD IUT VD 1DVTVD 1UT VD 1DD 1UT VD 1UT A If h Ax then x Alb AAx By Exercise 12a X is the orthogonal projection ofx onto RowA From part a and Exercise 12c Ax AAAx AAlAx Ax b LetAu b Since xl is the orthogonal projection of x onto Row A the Pythagorean Theorem shows that H u H2H xl H2 H u x Hz 2H xl H2 with equality only if u x 14 The leastsquares solutions of Ax b are precisely the solutions of Ax b where his the orthogonal projection of b onto ColA From Exercise 13 the minimum length solution ofo b is All so All is the minimum length leastsquares solution ofo b However b AAb by Exercise 12a and hence Ab AAAb Alb by Exercise 12c Thus Ab is the minimum length leastsquares solution ofo b 15 M The reduced SVD ofA is A UDVT where So the pseudoinverse Al VD71UT Ax b 966641 253758 034804 984443 0 0 185205 786338 589382 U D 0 262466 0 125107 398296 570709 0 0 109467 125107 398296 570709 3r3388 009549 633795 3r3388 009549 633795 andV 633380 023005 313529 633380 023005 313529 035148 999379 002322 may be calculated as well as the solution x Ab for the system H Chapter 7 Supplementary Exercises 423 05 35 325 325 05 35 325 325 A 05 15 175 175 2 8 05 15 175 175 8 10 30 150 150 6 Row reducing the augmented matrix for the system ATz 22 shows that this system has a solution so 22 0 1 0 1 is in Col AT Row A Abasis for NulA is 211212 1 0 and an arbitrary element of NulA is 1 0 0 0 u 0211 daz One computes that H 22 V13150 while H xu H13150 202 2d2 Thus if u 0 22 H lt 22 U H which con rms that 22 is the minimum length solution to Ax b M The reduced SVD of A is A UD V where 337977 591763 U 23 1428 694283 936307 290230 062526 187578 095396 129536 0 0 752053 D 0 144553 0 206232 0 0 337763 618696 690099 721920 050939 0 0 0 341800 387156 856320 637916 573534 513928 0 0 0 andV So the pseudoinverse A VrDilUrf may be calculated as well as the solution x Ab for the systemAx b 5 0 05 15 23 0 0 0 0 0 A 0 2 5 15 2 50 5 1 35 105 9 0 0 0 0 0 Row reducing the augmented matrix for the system ATz 5 shows that this system has a solution so 22 0 0 1 0 is in ColAT Row A Abasis for NulA is 211212 0 0 and an arbitrary element ofNulA is 0 0 0 1 ucalda2 One computes that llxlV31110 whileHxuH31110c2 d2 Thus ifu 0 H 22 H lt 32 u H which confirms that 22 is the minimum length solution to Ax b Vector Spaces 41 SOLUTIONS Notes This section is designed to avoid the standard exercises in which a student is asked to check ten axioms on an array of sets Theorem 1 provides the main homework tool in this section for showing that a set is a subspace Students should be taught how to check the closure axioms The exercises in this section and the next few sections emphasize R to give students time to absorb the abstract concepts Other vectors do appear later in the chapter the space S of signals is used in Section 48 and the spaces R of polynomials are used in many sections of Chapters 4 and 6 1 a If u and v are in V then their entries are nonnegative Since a sum of nonnegative numbers is nonnegative the vector u v has nonnegative entries Thus u v is in V 6quot 2 Example If u 2 and c 1 then u is in Vbut cu is not in V x x cx 2 a If u J is in W then the vector cu c J is in Wbecause cxcy c2 xy 2 0 y y 0 since xy 2 0 6quot 1 2 Example If u 7 and v 3 then u and v are in Wbut u v is not in W 5 3 Example If u J and c 4 then u is in H but cu is not in H Since H is not closed under scalar multiplication H is not a subspace of R2 4 Note that u and v are on the line L but u v is not a r u 11V 477 r 5 Yes Since the set is Span t2 the set is a subspace by Theorem 1 185 186 CHAPTER4 Vector Spaces 6 No The zero vector is not in the set 7 No The set is not closed under multiplication by scalars which are not integers 8 Yes The zero vector is in the set H If p and q are in H then p q0 p0 q0 0 0 0 so p q is in H For any scalar c cp0 C p0 c 0 0 so cp is inH Thus His a subspace by Theorem 1 l 9 The setH Span v where v 3 Thus His a subspace oflR3 by Theorem 1 2 2 10 The setH Span v where v 0 Thus His a subspace ofR3 by Theorem 1 l 5 2 11 The set W Spanu v where u l and v 0 Thus Wis a subspace oflR3 by Theorem 1 0 l l 3 l 1 4 12 The set W Spanu v where u 2 and v 1 Thus Wis a subspace ofR by Theorem 1 0 4 13 a The vector w is not in the set v1v2v3 There are 3 vectors in the set v1v2v3 Uquot The set Spanv1v2v3 contains in nitely many vectors P The vector w is in the subspace spanned by v1 v2 v3 if and only if the equation xlv1 xzv2 x3v3 w has a solution Row reducing the augmented matrix for this system of linear equations gives 1 2 4 3 l 0 0 l 0 l 2 1 0 l 2 l l 3 6 2 0 0 0 0 a so the equation has a solution and w is in the subspace spanned by v1v2v3 p A A The augmented matrix is found as in Exercise l3c Since 1 2 4 8 l 0 0 0 0 l 2 4 0 l 2 0 l 3 6 7 0 0 0 l the equation xlv1 xzv2 x3v3 w has no solution and w is not in the subspace spanned by V1 V2 9 V3 15 Since the zero vector is not in W W is not a vector space 16 Since the zero vector is not in W W is not a vector space 41 Solutions 187 17 Since a vector w in Wmay be written as 1 1 0 0 1 1 wa b 0 1 0 0 1 0 1 0 0 1 1 5 1 0 0 1 0 is a set that spans W 18 Since a vector w in Wmay be written as 4 3 0 0 0 0 wa b 0 1 1 1 2 0 1 4 3 0 0 0 0 S 1 1 2 0 1 is a set that spans W gt0 Let H be the set of all functions described by yt clcos wt czsin wt Then H is a subset of the vector space V of all realvalued functions and may be written as H Span cos wt sin wt By Theorem 1 H is a subspace of Vand is hence a vector space 20 a The following facts about continuous functions must be shown 1 The constant function ft 0 is continuous 2 The sum of two continuous functions is continuous 3 A constant multiple of a continuous function is continuous b LetH fin Ca b fa fb 1 Let gt 0 for all tin 1 b Then ga gb 0 so g is in H 2 Let g and h be inH Then ga gb and ha hb and g ha ga ha gb hb g hb so g h is in H 3 Let g be in H Then ga gb and cga cga cgb cgb so cg is in H Thus H is a subspace of Ca b 21 The setH is a subspace of M k2 The zero matrix is in H the sum of two upper triangular matrices is upper triangular and a scalar multiple of an upper triangular matrix is upper triangular 22 The setHis a subspace of ML The 2 x 4 zero matrix 0 is ingecause F0 0 IfA and B are matrices inH thenFA BFA FB0 0 0 soA Bis inH IfA is inHandc is ascalar then FcA CFA CO 0 so CA is inH 188 24 26 27 28 CHAPTER4 Vector Spaces 21 False The zero vector in Vis the function f whose values ft are zero for all t in R b False An arrow in threedimensional space is an example of a vector but not every arrow is a vector 0 False See Exercises 1 2 and 3 for examples of subsets which contain the zero vector but are not subspaces Q True See the paragraph before Example 6 D False Digital signals are used See Example 3 True See the de nition of a vector space True See statement 3 in the box before Example 1 True See the paragraph before Example 6 False See Example 8 False The second and third parts of the conditions are stated incorrectly For example part ii does not state that u and v represent all possible elements of H mPnc 24 213 b5 4 9072 JRLIIWOQ 95907 JkuquJk Consider u ilu By Axiom 10 u ilu lu ilu By Axiom 8 lu ilu l 71u Ou By Exercise 27 OH 0 Thus u ilu 0 and by Exercise 26 ilu iu By Axiom 10 u lu Since 0 is nonzero 0710 l and u 0 10u By Axiom 9 0 10u 071cu 0710 since cu 0 Thus u 0710 0 by Property 2 proven in Exercise 28 Any subspace H that contains u and v must also contain all scalar multiples of u and v and hence must also contain all sums of scalar multiples of u and v Thus H must contain all linear combinations of u and v or Spanu v Note Exercises 3234 provide good practice for mathematics majors because these arguments involve simple symbol manipulation typical of mathematical proofs Most students outside mathematics might pro t more from other types of exercises 32 Both H and K contain the zero vector of Vbecause they are subspaces of V Thus the zero vector of V is inH K Let u and v be inH K Then u and v are inH SinceHis a subspace u v is in H Likewise u andv are inK SinceKis a subspace u v is inK Thus u v is inH K Letu be inH K Then u is in H Since H is a subspace cu is in H Likewise v is inK Since Kis a subspace all is in K Thus cu is inH Kfor any scalar c and Hm Kis a subspace of V 33 43 A A proofthat H K Spanu1 41 Solutions 189 The union of two subspaces is not in general a subspace For an example in R2 let H be the xaxis and let K be the yaxis Then both H and K are subspaces of R2 but H U K is not closed under vector addition The subsetH U K is thus not a subspace of 1R2 21 Given subspaces H and K of a vector space V the zero vector of Vbelongs to H K because 0 is in bothH and Ksince they are subspaces and 0 0 0 Next take two vectors in H K say w1u1v1 and w2 u2 v2 where 111 and 112 are inH and v1 and v2 are inK Then w1w2 u1v1u2v2 u1u2v1v2 because vector addition in V is commutative and associative Now 111 112 is in H and v1 v2 is in K because H and K are subspaces This shows that w1 w2 is in H K Thus H K is closed under addition of vectors Finally for any scalar 0 CW1 Cll1V1 cu1 cv1 The vector cu1 belongs to H and cv1 belongs to K because H and K are subspaces Thus cw1 belongs to H K so H K is closed under multiplication by scalars These arguments show that H K satis es all three conditions necessary to be a subspace of V P Certainly H is a subset of H K because every vector u in H may be written as u 0 where the zero vector 0 is inK and also in H of course Since H contains the zero vector of H K andH is closed under vector addition and multiplication by scalars because H is a subspace of V H is a subspace of HK The same argument applies whenHis replaced by K so Kis also a subspace ofH K upv1vq has two parts First one must show thatH K is a subset of Spanu1u of H K 1 A typical vector H has the form clu1 cpup and a typical vector in Khas the form pv1vq Second one must show that Spanu1uPv1vq is a subset d1v1 dqvq The sum ofthese two vectors is a linear combination of u1upv1vq and so belongs to Spanu1upvlvq Thus HKis a subset of Spanu1upv1vq 2 Each of the vectors u1 combination of these vectors belongs to H K since H K is a subspace by Exercise 33a Thus v1vq belongs to H K by Exercise 33b and so any linear up Spanu1upv1vq is a subset ofHK M Since 7 4 9 9 l 0 0 152 4 5 4 7 0 l 0 3 2 1 4 4 N 0 0 1 112 9 7 7 8 0 0 0 0 w is in the subspace spanned by v1v2v3 M Since 5 5 9 l 0 0 ll 2 8 8 6 0 l 0 2 A y 5 9 3 l 0 0 l 7 2 3 2 7 4 0 0 0 0 y is in the subspace spanned by the columns ofA 190 CHAPTER4 Vector Spaces 37 M The graph of ft is given below A conjecture is that ft cos 4t The graph of ht is given below A conjecture is that ht sin 51 42 Solutions 191 42 SOLUTIONS Notes This section provides a review of Chapter 1 using the new terminology Linear tranforrnations are introduced quickly since students are already comfortable with the idea from Rquot The key exercises are 17726 which are straightforward but help to solidify the notions of null spaces and column spaces Exercises 30736 deal with the kernel and range of a linear transformation and are progressively more advanced theoretically The idea in Exercises 7714 is for the student to use Theorems 1 2 or 3 to determine whether a given set is a subspace 1 One calculates that 3 5 3 1 0 Aw6 2 0 30 8 4 1 4 0 so w is inNulA 2 One calculates that 5 21 19 5 0 Aw 13 23 2 3 0 8 14 1 2 0 so w is inNulA S43 First find the general solution ofo 0 in terms of the free variables Since 1 0 7 6 0 0 1 4 2 0 the general solution is x1 7x3 6x4 x2 4x3 2x4 with g and x4 free So x1 7 6 x x2 x3 4 x4 2 x3 1 0 x4 0 l and a spanning set for NulA is 7 6 4 2 1 0 0 1 4 First find the general solution ofo 0 in terms of the free variables Since 1 6 0 0 0 A 01 0 0 1 0 0 the general solution is x1 6x2 g 0 with x2 and x4 free So x1 6 0 x2 1 0 x x2 x4 x3 0 0 x4 0 l 192 CHAPTER4 Vector Spaces and a spanning set for NulA is 6 0 1 0 0 01 5 First nd the general solution ofo 0 in terms of the free variables Since 1 2 0 4 0 0 A 0 0 0 1 9 0 0 0 0 0 0 1 0 the general solution is x1 2x2 4x4 x3 9x4 x5 0 with x2 and x4 free So x1 2 4 x2 1 0 x x3 x2 0 x4 9 x4 0 1 x5 0 0 and a spanning set for NulA is 2 4 l 0 0 9 0 1 0 0 6 First nd the general solution ofo 0 in terms of the free variables Since 1 0 6 8 1 0 A 0 0 1 2 1 0 0 0 0 0 0 0 0 the general solutlon 1s x1 6x3 8x4 x5 x2 2x3 x4 w1th x3 x4 and x5 free So xli 67 8 1 x2 2 l 0 x x3 x3 lx4 0 x5 0 x4 0 1 0 x57 07 0 1 and a spanning set for NulA is 6 7 8 17 2 l 0 1 0 0 0 1 0 0 7 0 17 H H H H H gt1 9 gt0 42 Solutions 193 The set W is a subset of R3 If Wwere a vector space under the standard operations in R3 then it would be a subspace of R3 But Wis not a subspace of R3 since the zero vector is not in W Thus Wis not a vector space The set W is a subset of R3 If Wwere a vector space under the standard operations in R3 then it would be a subspace of R3 But Wis not a subspace of R3 since the zero vector is not in W Thus Wis not a vector space The set W is the set of all solutions to the homogeneous system of equations a 7 2b 7 4c 0 l 2 74 0 2a7c73d0 Thus WNulAwhere A2 0 1 3 Thus Wis a subspace oflR4 by Theorem 2 and is a vector space The set Wis the set of all solutions to the homogeneous system of equations a 3b 7 c 0 1 1 1 1 abc7d0ThusWNulAwhereA 1 0 Thus Wis a subspace of R4 by Theorem 2 and is a vector space The set W is a subset of R4 If Wwere a vector space under the standard operations in R4 then it would be a subspace of R4 But Wis not a subspace of R4 since the zero vector is not in W Thus Wis not a vector space The set W is a subset of R4 If Wwere a vector space under the standard operations in R4 then it would be a subspace of R4 But Wis not a subspace of R4 since the zero vector is not in W Thus Wis not a vector space An element w on Wmay be written as l 76 1 76 c 1 76 w c 0 d l 0 l 1 0 1 0 where c and d are any real numbers So W ColA where A 0 l Thus Wis a subspace of R3 by 1 0 Theorem 3 and is a vector space An element w on Wmay be written as 1 2 1 2 1 b 2 1 2 3 76 3 76 wa til where a and b are any real numbers So W ColA where A 1 2 1 2 Thus Wis a subspace oflR3 by 3 76 Theorem 3 and is a vector space 194 CHAPTER4 15 An element in this set may be written as 9 p A l 1 co 1 0 Vector Spaces 0 2 3 0 2 1 1 2 1 1 r S l 4 1 0 4 1 3 1 1 3 1 where r s and t are any real numbers So the set is ColA where A An element in this set may be written as l 0 l l 2 1 l 2 l b c d 0 5 4 0 5 0 1 0 0 where b c and d are any real numbers So the set is ColA where A The matrixA is a 4 X 2 matrix Thus a NulA is a subspace of R2 and b ColA is a subspace oflRA The matrixA is a 4 X 3 matrix Thus a NulA is a subspace of R3 and b ColA is a subspace oflRA The matrixA is a 2 X 5 matrix Thus a NulA is a subspace of R5 and b ColA is a subspace oflRZ The matrixA is a l X 5 matrix Thus a NulA is a subspace of R5 and b ColA is a subspace oflR1 R 30 1 0 0 0 A 0 0 0 0 0 0 0 kal no 1 2 0 0 Either column of A is a nonzero vector in ColA To nd a nonzero vector in NulA nd the general solution ofo 0 in terms of the free variables Since 42 Solutions 195 the general solution is x1 3x2 with x2 free Letting x2 be a nonzero value say x2 1 gives the nonzero vector xiiiHil which is in NulA Any column of A is a nonzero vector in ColA To nd a nonzero vector in NulA nd the general solution ofo 0 in terms of the free variables Since 1 0 7 6 0 A 0 5 2 0 0 1 4 the general solution is x1 7x3 6x4 x2 4x3 2x4 with g and x4 free Letting g and x4 be nonzero values say xj x4 1 gives the nonzero vector x1 1 x2 2 x x3 1 x4 1 which is in NulA Consider the system with augmented matrix A w Since 1 2 1 3 A w 0 0 0 the system is consistent and w is in ColA Also since 6 12 2 0 Aw 3 6 1 0 Wis inNulA Consider the system with augmented matrix A w Since 1 0 1 1 2 A w 0 1 1 2 1 0 0 0 0 the system is consistent and w is in ColA Also since 8 2 9 2 0 Aw 6 4 8 1 0 4 0 4 2 0 Wis inNulA a True See the de nition before Example 1 b False See Theorem 2 0 True See the remark just before Example 4 d False The equationAx b must be consistent for every b See 7 in the table on page 226 e True See Figure 2 f True See the remark after Theorem 3 196 CHAPTER 4 Vector Spaces 26 True See Theorem 2 True See Theorem 3 False See the box after Theorem 3 True See the paragraph after the de nition of a linear transformation True See Figure 2 H959 True See the paragraph before Example 8 27 LetA be the coefficient matrix of the given homogeneous system of equations SinceAx 0 for x 2 x is in NulA Since NulA is a subspace of R3 it is closed under scalar multiplication Thus 1 30 10x 20 is also in NulA and x1 30 x2 20 g 10 is also a solution to the system of 10 equations 28 LetA be the coefficient matrix of the given systems of equations Since the first system has a solution 0 the constant vector b l is in ColA Since ColA is a subspace of R3 it is closed under scalar 9 0 multiplication Thus 5b 5 is also in ColA and the second system of equations must thus have a 45 solution 29 21 Since A0 0 the zero vector is in ColA b Since AxAw AxwAxAw is in ColA 0 Since CAX Acxch is in ColA 30 Since T 0V 0W the zero vector 0W of W is in the range of T Let T x and T w be typical elements in the range of T Then since Tx Tw Tx wTxTw is in the range ofTand the range ofTis closed under vector addition Let c be any scalar Then since cTx T cx cTx is in the range of T and the range of T is closed under scalar multiplication Hence the range of T is a subspace of W 31 a Let p and q be arbitary polynomials in P2 and let 0 be any scalar Then pq0 p0q0 p0 q0 T T T pm an owl ip1q1l ip1liq1l 0 q and CPXO 100 T 6quot wood Cipa so T is a linear transformation 0Tp 33 42 Solutions 197 b Any quadratic polynomial q for which q0 0 and ql 0 will be in the kernel of T The x1 polynomial q must then be a multiple of pt tt 1 Given any vector x2 in R2 the polynomial p x1 x2 qt has p0 x1 and pl x2 Thus the range ofT is all oflRZ Any quadratic polynomial q for which q0 0 will be in the kernel of T The polynomial q must then be q at btz Thus the polynomials p1t t and p2 t t2 span the kernel of T If a vector is in the range of T it must be of the form a If a vector is of this form it is the image of the polynomial a pt a in P2 Thus the range ofT is a a real a a For anyA and B in M M2 and for any scalar c TABABABT ABAT 152T AATBBTTATB and TcA cAT CAT cTA so T is a linear transformation LetB be an element of M2X2 with BT B and let A B Then BlBTlBlBB 2 2 2 Uquot l l l T A AAT B B T 2 2 2 Part b showed that the range of T contains the set of all B in M 2X2 with BT B It must also be shown that any B in the range of T has this property Let B be in the range of T Then B T A for someA in MZXZ Then BAAT and P BT AATT AT ATT AT AAAT B soBhas the property that BT B 9 b LetAa d beinthekemelofT Then TAAAT0so c bac2a cb00 d b dbc 2d 0 0 Solving it is found that a d 0 and c b Thus the kernel ofTis ll aiml AATa C Let f and g be any elements in C0 l and let 0 be any scalar Then T t is the antiderivative F of f with F0 0 and T g is the antiderivative G of g with G0 0 By the rules for antidifferentiation F G will be an antiderivative of f g and F G0 F0 G0 0 0 0 Thus Tf g TD Tg Likewise CF will be an antiderivative of cf and cF0 cF0 CO 0 Thus Tcf cTf and T is a linear transformation To nd the kernel of T we must find all functions fin C 0 l with antiderivative equal to the zero function The only function with this property is the zero function 0 so the kernel of T is 0 198 CHAPTER4 Vector Spaces 35 Since U is a subspace of V 0V is in U Since T is linear T0V 0W So 0W is in TUB Let Tx and Ty be typical elements in TUB Then x and y are in U and since Uis a subspace of V x y is also in U Since Tis linear TxTy Txy So TxTy is in TUB and THO is closed under vector addition Let c be any scalar Then since x is in U and U is a subspace of V ex is in U Since T is linear T ex cTx and cTx is in T U Thus T 0 is closed under scalar multiplication and T 0 is a subspace of W 36 Since Z is a subspace of W 0W is in Z Since Tis linear T0V 0W So 0V is in U Let x and y be typical elements in U Then Tx and Ty are in Z and since Z is a subspace of W Tx Ty is also in Z Since Tis linear TxTy Txy So Txy is in Z and xy is in U Thus Uis closed under vector addition Let c be any scalar Then since x is in U T x is in Z Since Z is a subspace of W cTx is also in Z Since T is linear cTx Tcx and Tcx is in TUB Thus ax is in Uand U is closed under scalar multiplication Hence U is a subspace of V 43 l M Consider the system with augmented matrix A w Since 1 0 0 195 195 0 1 0 3919 2019 0 0 1 26795 17295 0 0 0 0 0 A Wl the system is consistent and w is in ColA Also since 7 6 4 1 l 14 w is not inNulA 38 M Consider the system with augmented matrix A w Since 1 0 l 0 2 0 l 2 0 3 0 0 0 l l 5 0 0 0 0 0 A w1 the system is consistent and w is in ColA Also since 8 5 2 0 LII N N Ob Nt 0 0 0 0 42 Solutions 199 39 M a To show that 213 and as are in the column space of B we can row reduce the matrices B 213 and B 213 1 0 0 1 3 0 1 0 1 3 B 33 N 0 0 1 0 0 0 0 0 1 0 0 10 3 0 1 0 26 3 B as N 0 0 1 4 0 0 0 0 Since both these systems are consistent a3 and as are in the column space of B Notice that the same conclusions can be drawn by observing the reduced row echelon form forA 1 0 1 3 0 10 3 0 1 1 3 0 26 3 N 0 0 0 1 4 0 0 0 0 0 A 9 We nd the general solution of Ax 0 in terms of the free variables by using the reduced row echelon form ofA given above x1 13g 103x5 x2 13x3 263x5 x4 4x5 with x3 and x5 free So x1 13 103 x2 13 263 x g x3 1 x5 0 x4 0 4 x5 0 l and a spanning set for NulA is 13 103 13 263 1 0 0 4 0 1 c The reduced row echelon form of A shows that the columns of A are linearly dependent and do not span R4 Thus by Theorem 12 in Section 19 T is neither onetoone nor onto 40 M Since the line lies both in H Spanv1v2 and in K Spanv3v4 w can be written both as 01v1 02v2 and c3v3 c4v4 To nd wwe must nd the cj s which solve 01v1 02v2 c3v3 c4v4 0 Row reduction of v1 v2 v3 v4 0 yields 51 200100 1030 33 ll20010 2630 8 4 5 28 0 0 0 l 4 0 200 43 CHAPTER4 Vector Spaces so the vector ofg s must be a multiple of103 7263 4 1 One simple choice is 10 726 12 3 which gives w 10v1 26v2 12v3 3v4 24 48 24 Another choice for w is 1 72 71 SOLUTIONS Notes The de nition for basis is given initially for subspaces because this emphasizes that the basis elements must be in the subspace Students often overlook this point when the de nition is given for a vector space see Exercise 25 The subsection on bases for Nul A and Col A is essential for Sections 45 and 46 The subsection on Two Views of a Basis is also fundamental to understanding the interplay between linearly independent sets spanning sets and bases Key exercises in this section are Exercises 21725 which help to deepen students understanding of these different subsets of a vector space 1 43 A 5quot Consider the matrix whose columns are the given set of vectors This 3 X 3 matrix is in echelon form and has 3 pivot positions Thus by the Invertible Matrix Theorem its columns are linearly independent and span R3 So the given set of vectors is a basis for R3 Since the zero vector is a member of the given set of vectors the set cannot be linearly independent and thus cannot be a basis for R3 Now consider the matrix whose columns are the given set of vectors This 3 X 3 matrix has only 2 pivot positions Thus by the Invertible Matrix Theorem its columns do not span R3 Consider the matrix whose columns are the given set of vectors The reduced echelon form of this matrix 1S 1 3 3 1 0 92 0 2 5 0 1 52 2 4 1 0 0 0 so the matrix has only two pivot positions Thus its columns do not form a basis for R3 the set of vectors is neither linearly independent nor does it span R3 Consider the matrix whose columns are the given set of vectors The reduced echelon form of this matrix is 2 1 7 1 0 0 2 3 5 0 1 0 1 2 4 0 0 1 so the matrix has three pivot positions Thus its columns form a basis for R3 Since the zero vector is a member of the given set of vectors the set cannot be linearly independent and thus cannot be a basis for R3 Now consider the matrix whose columns are the given set of vectors The reduced echelon form of this matrix is 1 2 0 0 1 0 0 0 3 9 0 3 0 1 0 0 0 0 0 5 0 0 0 1 so the matrix has a pivot in each row Thus the given set of vectors spans R3 43 Solutions 201 6 Consider the matrix whose columns are the given set of vectors Since the matrix cannot have a pivot in each row its columns cannot span R3 thus the given set of vectors is not a basis for R3 The reduced echelon form of the matrix is 1 4 1 0 2 5 0 1 3 6 0 0 so the matrix has a pivot in each column Thus the given set of vectors is linearly independent 7 Consider the matrix whose columns are the given set of vectors Since the matrix cannot have a pivot in each row its columns cannot span R3 thus the given set of vectors is not a basis for R3 The reduced echelon form of the matrix is 2 6 10 3 10 1 0 5 0 0 so the matrix has a pivot in each column Thus the given set of vectors is linearly independent 8 Consider the matrix whose columns are the given set of vectors Since the matrix cannot have a pivot in each column the set cannot be linearly independent and thus cannot be a basis for R3 The reduced echelon form of this matrix is 1 0 3 0 1 0 0 3 2 4 3 5 2010 12 3 1 4 2 0 0 1 1 2 so the matrix has a pivot in each row Thus the given set of vectors spans R3 9 We nd the general solution of Ax 0 in terms of the free variables by using the reduced echelon form of A l 0 3 2 1 0 3 2 0 1 5 4 0 1 5 4 3 2 1 2 0 0 0 0 So x1 3Jg 2x4 x2 5x3 4x4 with g and x4 free So x1 3 2 x2 5 4 x x3 x4 g l 0 x4 0 l and a basis for NulA is 202 CHAPTER4 Vector Spaces 10 We nd the general solution ofo 0 in terms of the free variables by using the reduced echelon form of A l 0 5 l 4 l 0 5 0 7 2 1 6 2 2 0 1 4 0 6 0 2 8 1 9 0 0 0 1 3 So x1 5x3 7x5 x2 4Jg 6x5 x4 3x5 with x3 and x5 free So x1 5 7 x2 4 6 x x3 x3 l x5 0 x4 0 x5 0 l and a basis for NulA is 5 7 4 6 1 0 0 3 0 11 Let A l 2 l Then we wish to nd a basis for NulA We nd the general solution ofo 0 in terms of the free variables x 72y 7 Z with y and Z free So 6 2 l x y y 1 2 0 Z 0 l andabasis for NulAis 2 1 1 0 0 l 12 We want to nd a basis for the set ofvectors in R2 in the line 5x 7y 0 Let A 5 l Then we wish to find a basis for NulA We nd the general solution ofo 0 in terms of the free variables y 5x with x free So man and a basis for NulA is ll 43 Solutions 203 13 Since B is a row echelon form of A we see that the first and second columns of A are its pivot columns Thus a basis for ColA is 2 4 2 6 3 8 To nd a basis for NulA we nd the general solution of Ax 0 in terms of the free variables x1 6x3 5x4 x2 52g 32x4 with 2 and x4 free So x1 6 5 x2 52 32 x x3 4 x3 1 0 x4 0 l and a basis for NulA is 6 5 5 2 32 1 0 0 1 14 Since B is a row echelon form of A we see that the first third and fth columns of A are its pivot columns Thus a basis for ColA is 1 5 3 2 5 2 1 0 5 3 5 2 To nd a basis for NulA we nd the general solution of Ax 0 in terms of the free variables mentally completing the row reduction of B to get x1 2x2 4x4 x3 7 5x4 x5 0 with x2 and x4 free x1 2 4 x2 1 0 x x3 x2 0 264 75 x4 0 1 x5 0 0 and a basis for NulA is 2 4 204 CHAPTER4 Vector Spaces 15 This problem is equivalent to nding a basis for ColA where A v1 v2 v3 v4 v5 Since the reduced echelon form ofA is 10 31210 304 01 4 3101 40 5 321 8 6 0001 2 2 367900000 we see that the first second and fourth columns of A are its pivot columns Thus a basis for the space spanned by the given vectors is 1 0 1 0 1 3 3 2 8 2 3 7 16 This problem is equivalent to nding a basis for ColA where A v1 v2 v3 v4 v5 Since the reduced echelon form of A is 1 2650100 1 2 01 1 33010 35 0 123 100102 11 1 4100000 we see that the first second and third columns of A are its pivot columns Thus a basis for the space spanned by the given vectors is 1 2 6 0 1 1 0 1 2 1 1 1 17 M This problem is equivalent to nding a basis for ColA where A v1 v2 v3 v4 v5 Since the reduced echelon form of A is 8 4 1 6 1 1 0 0 1 2 3 1 0 5 2 7 3 1 9 4 11 0 0 1 0 3 0 0 0 0 0 4 7 10 7 0 0 0 0 0 we see that the first second and third columns of A are its pivot columns Thus a basis for the space spanned by the given vectors is 8 4 1 9 5 4 3 1 9 6 4 6 18 p A D 22 Let A v1 Let A v1 43 Solutions 205 M This problem is equivalent to nding a basis for ColA where A v1 v2 v3 v4 v5 Since the reduced echelon form of A is 8 8 8 1 9 1 0 53 0 43 7 7 7 4 3 0 1 23 0 13 6 9 4 9 4 0 0 0 l 1 5 5 5 6 1 0 0 0 0 0 7 7 7 7 0 0 0 0 0 0 we see that the first second and fourth columns ofA are its pivot columns Thus a basis for the space spanned by the given vectors is 8 8 1 7 7 4 6 9 9 5 5 6 7 7 7 Since 4v1 5v2 3v3 0 we see that each of the vectors is a linear combination of the others Thus the sets v1v2 v1v3 and v2v3 all span H Since we may confirm that none of the three vectors is a multiple of any of the others the sets v1v2 v1v3 and v2v3 are linearly independent and thus each forms a basis for H Since v1 3v2 5v3 0 we see that each of the vectors is a linear combination of the others Thus the sets v1v2 v1v3 and v2v3 all span H Since we may confirm that none of the three vectors is a multiple of any of the others the sets v1v2 v1v3 and v2v3 are linearly independent and thus each forms a basis for H 21 False The zero vector by itself is linearly dependent See the paragraph preceding Theorem 4 b False The set b1bp must also be linearly independent See the definition ofa basis c True See Example 3 Q False See the subsection Two Views ofa Basis 5quot False See the box before Example 9 21 False The subspace spanned by the set must also coincide with H See the definition of a basis 6quot True Apply the Spanning Set Theorem to Vinstead of H The space Vis nonzero because the spanning set uses nonzero vectors True See the subsection Two Views of a Basis an False See the two paragraphs before Example 8 D False See the warning after Theorem 6 v2 v3 v4 ThenA is square and its colunms span R4 since R4 Spanv1v2v3v4 So its colunms are linearly independent by the Invertible lVIatrix Theorem and v1v2v3v4 is a basis for R4 vn ThenA is square and its colunms are linearly independent so its colunms span Rquot by the Invertible Matrix Theorem Thus v1 vn is a basis for Rquot 206 43 N LetA be the n X kmatrix v1 LetA be the n X kmatrix v1 CHAPTER4 Vector Spaces In order for the set to be a basis for H v1v2v3 must be a spanning set for H that is H Spanv1v2v3 The exercise shows thatH is a subset of Spanv1v2v3 but there are vectors in Spanv1v2v3 which are not inHv1 and v3 for example So H Spanv1v2v3 and v1v2v3 is not a basis for H Since sin t cos I 12 sin 21 the set sin t sin 21 spans the subspace By inspection we note that this set is linearly independent so sin t sin 21 is a basis for the subspace The set cos wt sin wt spans the subspace By inspection we note that this set is linearly independent so cos wt sin wt is a basis for the subspace The set 8 te bt spans the subspace By inspection we note that this set is linearly independent so 875 te bt is a basis for the subspace vk SinceA has fewer columns than rows there cannot be a pivot position in each row ofA By Theorem 4 in Section 14 the columns ofA do not span Rquot and thus are not a basis for Rquot v SinceA has fewer rows than columns rows there cannot be a pivot position in each column of A By Theorem 8 in Section 16 the columns of A are not linearly independent and thus are not a basis for Rquot Suppose that v1vp is linearly dependent Then there exist scalars 01cp not all zero with clvlcpvp 0 SinceT is linear Tclv1cpvp clTv1cPTvP and Tclv1cpvpT00 Thus clTv1cPTvP0 and since not all of the cl are zero T v1 T v P is linearly dependent Suppose that T v1 T v P is linearly dependent Then there exist scalars 01cp not all zero with clTv1cPTvP0 SinceT is linear Tclv1cpvpclTv1cPTvP0T0 SinceT is onetoone Tclv1cpvpT0 implies that clv1cPvP 0 Since not all of the cl are zero v1 VP is linearly dependent 43 Solutions 207 Neither polynomial is a multiple of the other polynomial So p1 p2 is a linearly independent set in P3 Note p1p2 is also a linearly independent set in P2 since p1 and p2 both happen to be in lP z By inspection p3 p1 p2 or p1 p2 p3 0 By the Spanning Set Theorem Spanp1p2p3 Spanplp2 Since neither p1 nor p2 is a multiple of the other they are linearly independent and hence p1p2 is a basis for Spanp1p2p3 Let v1v3 be any linearly independent set in avector space V and let v2 and v4 each be linear combinations of v1 and v3 For instance let v2 5v1 and v4 v1 v3 Then v1v3 is a basis for Spanv1avzav3av4 M Row reduce the following matrices to identify their pivot columns 1 0 2 l 0 2 2 2 2 0 l l h f H 111 112 113 3 1 7 0 0 so u1u2 1sa as1s or l l 3 0 0 0 l 2 17 l 0 3 0 2 4 0 l 2 v1 v2 v3 8 9 6 0 0 0so v1v2isabasisforK 4 5 27 0 0 0 71 0 2 l 2 l 2 2 2 0 2 4 u u u v v v l 2 3 1 2 3 3 1 7 8 9 6 1 1 3 4 5 2 71 0 2 0 2 4 0 l l 0 3 6 so u1u2vl 1s abas1s forHK 0 0 0 l 0 3 70 0 0 0 0 M For example writing cl itcz sin tc3cos 2tc4sin tcos t 0 with t 0 l 2 3 gives the following coef cent matriXA for the homogeneous systemAc 0 to four decimal places 0 sin 0 cos 0 sin 0 cos 0 0 0 l 0 1 sin 1 cos 2 sin 1 cos 1 l 0998 9801 0993 A 2 sin2 c0s4 sin2c0s2 2 1987 9211 194739 3 sin 3 cos 6 sin 3 cos 3 3 2955 8253 2823 This matrix is invertible so the systemAc 0 has only the trivial solution and t sin t cos 2t sin t cos I is a linearly independent set of functions 208 CHAPTER4 Vector Spaces 38 M For example writing 01 1c2 costc3 Acos2tc4 Acos3t05 Acos4t06 cos5tc7 cos6t 0 with t 0 1 2 3 4 5 6 gives the following coefficent matrixA for the homogeneous systemAc 0 to four decimal places 1 cos0 cos20 cos30 cos40 cos50 cos60 1 cos1 cos21 cos31 cos41 cos51 cos61 1 cos2 cos22 cos32 cos42 cos52 cos62 A1 cos3 cos23 cos33 cos43 cos53 cos63 1 cos4 cos24 cos34 cos44 cos54 cos64 1 cos5 cos25 cos35 cos45 cos55 cos65 71 cos6 cos26 cos36 cos46 cos56 cos66 71 1 1 1 1 1 1 1 9950 9900 9851 9802 9753 9704 1 9801 9605 9414 9226 9042 8862 1 9553 9127 8719 8330 7958 7602 1 9211 8484 7814 7197 6629 6106 1 8776 7702 6759 5931 5205 4568 71 8253 6812 5622 4640 3830 3161 This matrix is invertible so the system Ac 0 has only the trivial solution and 1 cos I coszt cos3t cos4t cosst cos t is a linearly independent set of functions 44 SOLUTIONS Notes Section 47 depends heavily on this section as does Section 54 It is possible to cover the Rquot parts of the two later sections however if the rst half of Section 44 and perhaps Example 7 is covered The linearity of the coordinate mapping is used in Section 54 to find the matrix of a transformation relative to two bases The changeofcoordinates matrix appears in Section 54 Theorem 8 and Exercise 27 The concept of an isomorphism is needed in the proof of Theorem 17 in Section 48 Exercise 25 is used in Section 47 to show that the changeofcoordinates matrix is invertible 1 We calculate that 3 4 3 x 5 3 H l 61 71 2 We calculate that x8lillt5gtlll 3 We calculate that 1 5 4 1 x3 4 0 2 1 7 5 3 2 0 9 A II l on gt0 9 N S43 44 Solutions 209 We calculate that l 3 4 0 x 4 2 8 5 7 7 1 0 2 3 5 1 0 8 The matr1x b1 b2 x row reduces to so XB 0 l 5 5 1 0 6 The matr1x b1 b2 x row reduces to 0 1 2 so xB 2 l 0 0 1 1 The matrix b1 b2 b3 x row reduces to 0 l 0 l so xB l 0 l 3 3 0 0 2 2 Thematrix b1 b2 b3 x row reduces to 0 l 0 0 so XB 0 0 0 l 5 5 The changeofcoordinates matrix from B to the standard basis in R2 is b1 b22 1 P B 98 The changeofcoordinates matrix from B to the standard basis in R3 is 3 2 8 PBb1 b2 b3 1 0 2 4 5 7 Since P871 converts x into its Bcoordinate vector we nd that l57 5254 Since P871 converts x into its Bcoordinate vector we nd that H 131 4 6 12 72 3 2 7 x x B B 5 7 0 52 2 0 5 We must nd cl czand 03 such that 011t202tt2c312tt2ptl4t7t2 Equating the coefficients of the two polynomials produces the system of equations cl c3 l 02 203 4 cl 02 c3 7 210 CHAPTER 4 4 H H p A l p A o0 gt0 1 We must solve the vector equation x1 3 x2 Vector Spaces We row reduce the augmented matrix for the system of equations to nd 1011 10 0 2 2 0 1 2 40 10 6sopB 6 1117 0 01 1 1 One may also solve this problem using the coordinate vectors of the given polynomials relative to the standard basis 1 1 t2 the same system of linear equations results We must nd cl CZ and 03 such that c11 02t t2 032 2tt2 pt 3t 6t2 Equating the coefficients of the two polynomials produces the system of equations cl 203 3 02 263 1 cl 02 c3 6 We row reduce the augmented matrix for the system of equations to nd 10 2 3 100 7 7 0 1 2 1010 3sopB 3 1 1 1 6 001 2 2 One may also solve this problem using the coordinate vectors of the given polynomials relative to the standard basis 1 1 t2 the same system of linear equations results a True See the de nition of the Bcoordinate vector b False See Equation 4 0 False P3 is isomorphic to R4 See Example 5 a True See Example 2 b False By de nition the coordinate mapping goes in the opposite direction 0 True If the plane passes through the origin as in Example 7 the plane is isomorphic to R2 2 3 1 8 2 7 I We row reduce the augmented matrix for the system of equations to nd 1 2 3 l 0 5 i3 8 7 ilNiO 1 1 Thus we can let x1 5 5x3 and x2 il 2 2g where x3 can be any real number Letting x3 0 and 1 x3 1 produces two different ways to express 1 as a linear combination of the other vectors 5v1 2v2 and 10v1 3v2 v3 There are infmtely many correct answers to this problem Foreachk bk0Ablml bkm0 bnso bkB 0l0ek The set S spans Vbecause every x in Vhas a representation as a unique linear combination of elements in S To show linear independence suppose that S v1 VH and that 01v1 cnvn 0 for some scalars cl an The case when cl On 0 is one possibility By hypothesis this is the unique 44 Solutions 211 and thus the only possible representation of the zero vector as a linear combination of the elements in S So S is linearly independent and is thus a basis for V For win Vthere exist scalars k1 k2 k3 and k4 such that 1 because v1v2v3v4 spans V Because the set is linearly dependent there exist scalars cl 02 c3 and w klv1 kzv2 k3V3 k4V4 04 not all zero such that 0clv1czv2 c3v3 c4v4 2 Adding l and 2 gives w w 0k1 clv1k2 czv2 k3 c3v3 k4 c4v4 3 At least one of the weights in 3 differs from the corresponding weight in 1 because at least one of the cl is nonzero So w is expressed in more than one way as a linear combination of v1 v2 v3 and v4 1 2 19 2 4 9 4 139 The matrix of the transformation will be P371 The matrix of the transformation will be P871 b1 b Tl Suppose that 51 Illa WB c 71 By de nition of coordinate vectors uwclb1mcnbw Since 11 and w were arbitrary elements of V the coordinate mapping is onetoone Given y y1yn in Rquot let u ylb1 m ynbn Then by definition uB y Since y was arbitrary the coordinate mapping is onto Rquot Since the coordinate mapping is onetoone the following equations have the same solutions 01 CF clu1m cpup 0 the zero vector in V 4 clu1 cpup 18 0B the zero vector in Rquot 5 Since the coordinate mapping is linear 5 is equivalent to 0 clu1BmcPuPB E 6 0 Thus 4 has only the trivial solution if and only if 6 has only the trivial solution It follows that u1up is linearly independent ifand only if u1BupB is linearly independent This result also follows directly from Exercises 31 and 32 in Section 43 212 CHAPTER4 Vector Spaces 26 By de nition w is a linear combination of u1u if and only if there exist scalars 01c such that P P wclu1mcpup 7 Since the coordinate mapping is linear WB Clu1B Cpup3 8 Conversely 8 implies 7 because the coordinate mapping is onetoone Thus w is a linear combination of u1up ifand only if WB is a linear combination of u1uP Note Students need to be urged to write not just to compute in Exercises 27734 The language in the Study Guide solution of Exercise 31 provides a model for the students In Exercise 32 students may have difficulty distinguishing between the two isomorphic vector spaces sometimes giving a vector in R3 as an answer for part b 27 The coordinate mapping produces the coordinate vectors 1 0 0 l 3 l 72 0 and 0 fl 3 71 respectively We test for linear independence of these vectors by writing them as columns of a matrix and row reducing 130100 01 1010 0 2300139 10 1000 Since the matrix has a pivot in each column its columns and thus the given polynomials are linearly independent 28 The coordinate mapping produces the coordinate vectors 1 0 72 73 0 l 0 l and l 3 72 0 respectively We test for linear independence of these vectors by writing them as columns of a matrix and row reducing 1 0 1 0 1 0 1 3 0 1 3 2 0 2 0 0 0 I 310000 Since the matrix does not have a pivot in each column its columns and thus the given polynomials are linearly dependent 29 The coordinate mapping produces the coordinate vectors 1 72 l 0 72 0 0 l and 8 l2 76 1 respectively We test for linear independence of these vectors by writing them as columns of a matrix and row reducing l 2 8 1 0 6 2 0 12 0 1 l 1 0 6 0 0 0 I 0 l 1 0 0 0 Since the matrix does not have a pivot in each column its columns and thus the given polynomials are linearly dependent 30 32 44 Solutions 213 The coordinate mapping produces the coordinate vectors 1 73 3 71 4 712 9 0 and 0 0 3 41 respectively We test for linear independence of these vectors by writing them as columns of a matrix and row reducing 1 4 0 10 4 3 12 0 01 1 3 9 300 039 1 0 4 00 0 Since the matrix does not have a pivot in each column its columns and thus the given polynomials are linearly dependent In each part place the coordinate vectors of the polynomials into the columns of a matrix and reduce the matrix to echelon form 1 3 4 1 1 3 4 1 a 3 5 5 0 0 4 7 3 5 7 6 1 0 0 0 0 Since there is not a pivot in each row the original four column vectors do not span R3 By the isomorphism between R3 and R the given set of polynomials does not span P2 0 1 3 2 1 2 2 0 5 8 4 3 0 2 6 3 1 2 2 0 0 0 0 7 2 9 Since there is a pivot in each row the original four column vectors span R3 By the isomorphism between R3 and 1 the given set of polynomials spans 11 a Place the coordinate vectors of the polynomials into the columns of a matrix and reduce the matrix to 1 2 1 1 2 1 echelon form 0 1 2 0 1 2 1 3 4 0 0 3 The resulting matrix is invertible since it row equivalent to 13 The original three column vectors form a basis for R3 by the Invertible Matrix Theorem By the isomorphism between R3 and E the corresponding polynomials form a basis for 1P2 b Since qB 3 1 2 q 3p1 p2 2p3 One might do the algebra in P2 or choose to compute 1 2 1 3 1 0 1 2 1 3 This combination of the columns of the matrix corresponds to the same 1 3 4 2 8 combination of p1 p2 and p3 So qt13t 8t2 The coordinate mapping produces the coordinate vectors 3 7 0 0 5 1 0 72 0 1 72 0 and 1 16 4 2 respectively To determine whether the set of polynomials is a basis for R we investigate whether the coordinate vectors form a basis for R4 Writing the vectors as the columns of a matrix and row reducing 35011002 71116 010 1 00 2 60013 0 2020000 214 CHAPTER4 Vector Spaces we nd that the matrix is not row equivalent to 14 Thus the coordinate vectors do not form a basis for R4 By the isomorphism between 1quot and P3 the given set of polynomials does not form a basis for P3 34 The coordinate mapping produces the coordinate vectors 5 73 4 2 9 l 8 76 6 72 5 0 and 0 0 0 1 respectively To determine whether the set of polynomials is a basis for R we investigate whether the coordinate vectors form a basis for R4 Writing the vectors as the columns of a matrix and row reducing 596010340 31 20 01140 48500001 2 60100 00 we nd that the matrix is not row equivalent to 14 Thus the coordinate vectors do not form a basis for R4 By the isomorphism between R4 and 11 the given set of polynomials does not form a basis for 11 35 To show that x is in H Spanv1v2 we must show that the vector equation xlv1 xzv2 x has a solution The augmented matrix v1 v2 x may be row reduced to show 11 14 19 1 0 5 3 5 8 13 0 1 8 3 10 13 18 0 0 0 I 7 10 15 0 0 0 Since this system has a solution x is in H The solution allows us to nd the Bcoordinate vector forx 5 3 since Xle1x2V2 53v183v2 XB 83 36 To show that x is in H Spanv1v2v3 we must show that the vector equation xlv1 xzv2 x3v3 x has a solution The augmented matrix v1 v2 v3 x may be row reduced to show 6 8 9 4 1 0 0 3 4 3 5 7 0 1 0 5 9 7 8 8 0 0 1 2 I 4 3330000 The first three columns show thatB is a basis for H Moreover since this system has a solution x is in H The solution allows us to nd the Bcoordinate vector for x since 3 xx1v1xzv2 x3v3 3v15v2 2v3 XB 5 2 12 26 0 0 37 We are given that XB 14 where B l5 3 0 To find the coordinates ofx relative 16 0 0 48 to the standard basis in R3 we must nd x We compute that 26 0 0 1 2 13 xPBxB 15 3 0 14 0 0 0 48 16 08 45 Solutions 215 1 2 26 0 0 38 We are given that XB 1 2 where B 15 3 0 To find the coordinates of x relative 1 3 0 0 48 to the standard basis in R3 we must find x We compute that 26 0 0 1 2 13 xPBxB 15 3 0 12 075 0 0 48 13 16 45 SOLUTIONS Notes Theorem 9 is true because a vector space isomorphic to Rquot has the same algebraic properties as Rquot a proof of this result may not be needed to convince the class The proof of Theorem 9 relies upon the fact that the coordinate mapping is a linear transformation which is Theorem 8 in Section 44 If you have skipped this result you can prove Theorem 9 as is done in Introduction to Linear Algebra by Serge Lang Springer Verlag New York 1986 There are two separate groups of truefalse questions in this section the second batch is more theoretical in nature Example 4 is useful to get students to visualize subspaces of different dimensions and to see the relationships between subspaces of different dimensions Exercises 31 and 32 investigate the relationship between the dimensions of the domain and the range of a linear transformation Exercise 32 is mentioned in the proof of Theorem 17 in Section 48 1 2 1 This subspace is H Spanv1v2 where v1 1 and v2 1 Since v1 and v2 are not multiples 0 3 of each other v1 v2 is linearly independent and is thus a basis for H Hence the dimension of H is 2 4 0 2 This subspace is H Spanltv1v2 where v1 3 and v2 0 Since v1 and v2 are not multiples 0 1 of each other v1 v2 is linearly independent and is thus a basis for H Hence the dimension of H is 2 0 0 2 1 3 Th1s subspace 1s HSpanv1v2v3 where v1 0 v2 1 and v3 3 Theorem4m 1 2 0 Section 43 can be used to show that this set is linearly independent v1 0 v2 is not a multiple of VI and since its first entry is not zero v3 is not a linear combination of v1 and v2 Thus v1v2v3 is linearly independent and is thus a basis for H Alternatively one can show that this set is linearly independent by row reducing the matrix v1 v2 v3 0 Hence the dimension of the subspace is 3 1 2 3 4 0 This subspace is H Spanv1v2 where v1 and v2 1 Since v1 and v2 are not multiples 0 1 of each other v1 v2 is linearly independent and is thus a basis for H Hence the dimension of H is 2 216 CHAPTER 4 I H H 5 9 gt1 9 0 Vector Spaces 4 2 5 4 Th1s subspace 1s HSpanv1V2V3 Where v1 1 v2 0 and v3 2 Smce v3 2v1 3 7 6 v1v2 v3 is linearly dependent By the Spanning Set Theorem v3 may be removed from the set with no change in the span ofthe set so H Spanv1v2 Since v1 and v2 are not multiples of each other v1 v2 is linearly independent and is thus a basis for H Hence the dimension of H is 2 3 6 1 2 Th1s subspace 1s HSpanv1v2v3 where v1 9 v2 5 and v3 3 Since 3 l 1 v3 l3v1 v1v2v3 is linearly dependent By the Spanning Set Theorem v3 may be removed from the set with no change in the span of the set so H Spanv1v2 Since v1 and v2 are not multiples of each other v1 v2 is linearly independent and is thus a basis for H Hence the dimension of H is 2 l 3 1 1 0 0 0 This subspace is HNulA where A 0 l 2 Since A 0 0 l 0 0 the 0 2 1 0 0 1 0 homogeneous system has only the trivial solution Thus H NulA 0 and the dimension of H is 0 From the equation a 7 3b c 0 it is seen that a b c d b3 l 0 0 C7l 0 l 0 d0 0 0 1 Thus the subspace is H Span v1v2v3 where v1 3100 v2 l0l0 and v3 00 01 It is easily checked that this set of vectors is linearly independent either by appealing to Theorem 4 in Section 43 or by row reducing v1 v2 v3 0 Hence the dimension of the subspace is 3 a 1 0 This subspace isH b abin R Spanv1v2 where v1 0 and v2 l Since v1 and a l 0 v2 are not multiples of each other v1 v2 is linearly independent and is thus a basis for H Hence the dimension of H is 2 The matrixA with these vectors as its columns row reduces to L 4 M 3 il 10 There are two pivot columns so the dimension of ColA which is the dimension of H is 2 The matrixA with these vectors as its columns row reduces to 1 3 9 7 1 0 3 2 0 1 4 3 0 1 4 3 2 1 2 1 0 0 0 0 There are two pivot columns so the dimension of ColA which is the dimension of the subspace spanned by the vectors is 2 p A N p A D 20 45 Solutions 217 The matrixA with these vectors as its columns row reduces to 1 3 8 3 1 0 7 0 2 4 6 0 0 1 5 0 0 1 5 7 0 0 0 1 There are three pivot columns so the dimension of ColA which is the dimension of the subspace spanned by the vectors is 3 The matrixA is in echelon form There are three pivot columns so the dimension of ColA is 3 There are two columns without pivots so the equationAx 0 has two free variables Thus the dimension of NulA is 2 The matrixA is in echelon form There are three pivot columns so the dimension of ColA is 3 There are three columns without pivots so the equationAx 0 has three free variables Thus the dimension of NulA is 3 The matrixA is in echelon form There are two pivot columns so the dimension of ColA is 2 There are two colunms without pivots so the equationAx 0 has two free variables Thus the dimension of NulA is 2 The matrixA row reduces to 1 3 13113 31 There are two pivot columns so the dimension of ColA is 2 There are no columns without pivots so the equationAx 0 has only the trivial solution 0 Thus NulA 0 and the dimension of NulA is 0 The matrixA is in echelon form There are three pivot columns so the dimension of ColA is 3 There are no colunms without pivots so the equationAx 0 has only the trivial solution 0 Thus NulA 0 and the dimension of Nul A is 0 The matrixA is in echelon form There are two pivot columns so the dimension of ColA is 2 There is one colunm without a pivot so the equationAx 0 has one free variable Thus the dimension of NulA is 1 a True See the box before Example 5 False The plane must pass through the origin see Example 4 False The dimension of Pquot is n 1 see Example 1 False The set S must also have n elements see Theorem 12 rDPnv True See Theorem 9 False The set R2 is not even a subset of R3 False The number of free variables is equal to the dimension of NulA see the box before Example 5 avg False A basis could still have only nitely many elements which would make the vector space finite dimensional Q False The set S must also have n elements see Theorem 12 D True See Example 4 218 CHAPTER 4 I 21 Vector Spaces The matrix whose columns are the coordinate vectors of the Hermite polynomials relative to the standard basis 1 t t2t3 of P3 is l 0 2 0 0 2 0 12 0 0 4 0 I 0 0 0 8 This matrix has 4 pivots so its columns are linearly independent Since their coordinate vectors form a linearly independent set the Hermite polynomials themselves are linearly independent in P3 Since there are four Hermite polynomials and dim 1P3 4 the Basis Theorem states that the Hermite polynomials form a basis for R A The matrix whose columns are the coordinate vectors of the Laguerre polynomials relative to the standard basis 11 t2t3 of P3 is 1 1 2 6 0 l 4 18 A 0 0 l 9 0 0 0 1 This matrix has 4 pivots so its columns are linearly independent Since their coordinate vectors form a linearly independent set the Laguerre polynomials themselves are linearly independent in P3 Since there are four Laguerre polynomials and dim 1P3 4 the Basis Theorem states that the Laguerre polynomials form a basis for P3 The coordinates of pt 7 12t 81 2 121 3 with respect to B satisfy 011 0221 c3 2 4t2 c4 12t 8t3 7 12t 8t2 123 Equating coefficients of like powers of 1 produces the system of equations cl 203 7 202 1204 12 403 8 804 12 3 Solving this system gives cl 3 02 3 c3 2 c4 32 and pB 3 3 2 The coordinates of pt 7 81 31 2 with respect to B satisfy cll02l tc32 4tt27 8t3t2 Equating coefficients of like powers of 1 produces the system of equations cl 02 203 7 cz 463 8 c3 3 5 Solving this system gives cl 5 02 4 c3 3 and pB 4 3 30 45 Solutions 219 Note rst that n 2 1 since S cannot have fewer than 1 vector Since n 2 l V 0 Suppose that S spans V and that S contains fewer than n vectors By the Spanning Set Theorem some subset S39 of S is a basis for V Since S contains fewer than n vectors and S39 is a subset of S S39 also contains fewer than n vectors Thus there is a basis S39 for Vwith fewer than n vectors but this is impossible by Theorem 10 since dimV n Thus S cannot span V IfdimV dimH 0 then V 0 andH 0 soH V Suppose thatdim V dimHgt 0 ThenH contains a basis S consisting of n vectors But applying the Basis Theorem to V S is also a basis for V Thus H V SpanS Suppose that dim lP k lt 00 Now lquot is a subspace of for all n and dim PM k so dim PH dim P This would imply that lP H P which is clearly untrue for example pt 1 6 is in l but not in PH Thus the dimension of l cannot be finite The space CR contains l as a subspace If CR were nitedimensional then P would also be finite dimensional by Theorem 11 Buth is in nitedimensional by Exercise 27 so CR must also be infinite dimensional a True Apply the Spanning Set Theorem to the set v1vp and produce a basis for V This basis will not have more than p elements in it so dimVS p Uquot True By Theorem 11 v1vp can be expanded to nd a basis for V This basis will have at least p elements in it so dimVZ p O True Take any basis which will contain p vectors for V and adjoin the zero vector to it 21 False For a counterexample let v be a nonzero vector in R3 and consider the set v 2v This is a linearly dependent set in R3 but dim R3 3 gt 2 b True If dimVS p there is a basis for Vwith p or fewer vectors This basis would be a spanning set for Vwith p or fewer vectors which contradicts the assumption False For a counterexample let v be a nonzero vector in R3 and consider the set v 2v This is a linearly dependent set in R3 with 3 7 l 2 vectors and dim R3 3 O Since H is a nonzero subspace of a finitedimensional vector space V H is finitedimensional and has a basis Let u1up be a basis for H We show that the set Tu1Tup spans TH Lety be in TH Then there is a vector x inHwith Tx y Since x is in Hand u1up is a basis for H x may be written as x clu1 cpup for some scalars 01cp Since the transformation Tis linear yTxTclu1cpup01Tu1CPTup Thus y is a linear combination of Tu1Tup and Tu1Tup spans TH By the Spanning Set Theorem this set contains a basis for T H This basis then has not more than p vectors and dimT H S p dim H Since H is a nonzero subspace of a finitedimensional vector space V H is finitedimensional and has a basis Let u1up be a basis for H In Exercise 31 above it was shown that Tu1 Tup spans T H In Exercise 32 in Section 43 it was shown that Tu1Tup is linearly independent Thus Tu1TuP is a basis for TH and dimTH p dim H 220 CHAPTER4 Vector Spaces 33 M a To nd a basis for R5 which contains the given vectors we row reduce 99610000100 1300137 74701000010000157 81 800100001 13000 37 565000100000103227 7 7 700001000 001 9 537 The rst second third fth and sixth columns are pivot columns so these columns of the original matrix v1v2v3e2e3 form a basis for R5 b The original vectors are the first k columns of A Since the set of original vectors is assumed to be linearly independent these columns of A will be pivot colunms and the original set of vectors will be included in the basis Since the columns of A include all the colunms of the identity matrix ColA Rquot 34 M a The Bcoordinate vectors of the vectors in C are the columns of the matrix 1 0 l 0 1 0 1 0 1 0 3 0 5 0 0 0 2 0 8 0 18 P 0 0 0 4 0 20 0 0 0 0 0 8 0 48 0 0 0 0 0 16 0 0 0 0 0 0 0 32 The matrix P is invertible because it is triangular with nonzero entries along its main diagonal Thus its colunms are linearly independent Since the coordinate mapping is an isomorphism this shows that the vectors in C are linearly independent b We know that dim H 7 because B is a basis for H Now C is a linearly independent set and the vectors in C lie in H by the trigonometric identities Thus by the Basis Theorem C is a basis for H 46 SOLUTIONS Notes This section puts together most of the ideas from Chapter 4 The Rank Theorem is the main result in this section Many students have difficulty with the difference in nding bases for the row space and the colunm space of a matrix The first process uses the nonzero rows of an echelon form of the matrix The second process uses the pivots columns of the original matrix which are usually found through row reduction Students may also have problems with the varied effects of row operations on the linear dependence relations among the rows and colunms of a matrix Problems of the type found in Exercises 19726 make excellent test questions Figure l and Example 4 prepare the way for Theorem 3 in Section 61 Exercises 27729 anticipate Example 6 in Section 74 46 Solutions 221 1 The matrix B is in echelon form There are two pivot columns so the dimension of ColA is 2 There are two pivot rows so the dimension of RowA is 2 There are two columns without pivots so the equation Ax 0 has two free variables Thus the dimension of NulA is 2 A basis for ColA is the pivot columns of A 1 4 1 2 5 6 A basis for Row A is the pivot rows ofB 10 l 50 2 5 6 To find a basis for NulA row reduce to reduced echelon form 1 0 1 5 A 0 1 5 2 3 The solution to Ax 0 in terms of free variables is x1 x3 5x4 x2 5 2x3 3x4 with x3 and x4 free Thus a basis for NulA is 1 5 52 3 1 0 0 1 2 The matrix B is in echelon form There are three pivot columns so the dimension of ColA is 3 There are three pivot rows so the dimension of Row A is 3 There are two columns without pivots so the equation Ax 0 has two free variables Thus the dimension of NulA is 2 A basis for ColA is the pivot columns of A 1 4 9 2 6 10 3 6 3 3 4 0 A basis for RowA is the pivot rows ofB 1 3 05 7 00 2 3 800005 To nd a basis for NulA row reduce to reduced echelon form 1 3 0 5 0 0 0 l 3 2 0 0 0 0 0 l I 0 0 0 0 0 The solution to Ax 0 in terms of free variables is x1 3x2 5x4 g 3 2x4 x5 0 with x2 and x4 free Thus a basis for NulA is 222 CHAPTER4 Vector Spaces 3 The matrix B is in echelon form There are three pivot columns so the dimension of ColA is 3 There are three pivot rows so the dimension of Row A is 3 There are two columns without pivots so the equation Ax 0 has two free variables Thus the dimension of NulA is 2 A basis for ColA is the pivot columns 2 6 2 2 3 3 4 9 5 2 3 4 A basis for RowA is the pivot rows ofB 2 36 250 03 ll 00 0 l3 To nd a basis for NulA row reduce to reduced echelon form 1 3 2 0 0 9 2 0 0 l 0 4 3 0 0 l 3 I 0 0 0 0 0 The solution to Ax 0 in terms of free variables is x1 3 2x2 9 2x5 2 43x5 x4 3x5 with x2 and x5 free Thus a basis for NulA is 32 92 0 0 43 0 3 0 1 4 The matrix B is in echelon form There are three pivot columns so the dimension of ColA is 3 There are three pivot rows so the dimension of Row A is 3 There are three columns without pivots so the equation Ax 0 has three free variables Thus the dimension of NulA is 3 A basis for ColA is the pivot columns of A 1 1 7 1 10 l l 1 l 3 5 1 2 0 A basis for Row A is the pivot rows of B 1 1 3 7 9 9 0 l 3 4 3 0 0 01 1 2 To nd a basis for NulA row reduce to reduced echelon form 1 0 2 0 9 2 1 0 0 1 0 0 0 0 0 0 0 0000 UI Ch gt1 9 gt0 9 H H p A N p A Ch H 46 Solutions 223 The solution to Ax 0 in terms of free variables is x1 2Jg 9x5 2x6 x2 x3 7x5 3x6 x4 x5 2x6 with x3 x5 and x6 free Thus a basis for NulA is 9 2 l 7 3 1 0 0 0 l 2 0 l 0 0 0 1 By the Rank Theorem dimNul A 8 rank A 8 3 5 Since dimRow A rank AdimRow A 3 Since rank AT dimColAT dimRowA rankAT 3 By the Rank Theorem dimNul A 3 rank A 3 3 0 Since dimRow A rank A dimRow A 3 Since rank AT dimCol AT dimRow A rank AT 3 Yes ColA R4 SinceA has four pivot columns dimCol A 4 Thus ColA is a fourdimensional subspace of R4 and ColA R4 No NulA R3 It is true that dimNul A 3 but NulA is a subspace oflR7 SinceA has four pivot columns rankA 4 and dimNul A 6 rankA 6 4 2 No ColA R4 It is true that dimColA rankA 4 but ColA is a subspace of R5 Since dimNul A 4 rankA 6 dimNulA 6 4 2 So dimColA rank A 2 Since dimNulA5rankA6 dimNulA6 5l So dimColArankAl Since dimNul A 2 rankA 5 dimNul A 5 2 3 So dimRow A dimColA rank A 3 Since dimNul A 4 rankA 6 dimNulA 6 4 2 So dimRow A dimColA rankA 2 The rank of a matrixA equals the number of pivot positions which the matrix has If A is either a 7X 5 matrix or a 5X 7 matrix the largest number of pivot positions thatA could have is 5 Thus the largest possible value for rankA is 5 The dimension of the row space of a matrixA is equal to rankA which equals the number of pivot positions which the matrix has IfA is either a 4X 3 matrix or a 3X 4 matrix the largest number of pivot positions thatA could have is 3 Thus the largest possible value for dimRow A is 3 Since the rank of A equals the number of pivot positions which the matrix has andA could have at most 6 pivot positions rank A S 6 Thus dimNul A 8 rankA 2 8 6 2 Since the rank of A equals the number of pivot positions which the matrix has andA could have at most 4 pivot positions rankA S 4 Thus dimNulA 4 rankA 2 4 4 0 a True The rows of A are identi ed with the columns of AT See the paragraph before Example 1 b False See the warning after Example 2 c True See the Rank Theorem d False See the Rank Theorem e True See the Numerical Note before the Practice Problem 224 p A 0 CHAPTER 4 Vector Spaces False Review the warning after Theorem 6 in Section 43 False See the warning after Example 2 True See the remark in the proof of the Rank Theorem True This fact was noted in the paragraph before Example 4 It also follows from the fact that the rows of AT are the columns of AT T A 9an e True See Theorem 13 Yes Consider the system as Ax 0 where A is a 5X6 matrix The problem states that dimNulA 1 By the Rank Theorem rankA 6 dimNulA 5 Thus dim ColA rank A 5 and since ColA is a subspace of R5 ColA R5 So every vector b in R5 is also in ColA and Ax b has a solution for all h No Consider the system as Ax b where A is a 6X8 matrix The problem states that dimNulA 2 By the Rank Theorem rank A 8 dimNul A 6 Thus dimCol A rank A 6 and since ColA is a subspace of R6 ColA R6 So every vector b in R6 is also in ColA and Ax b has a solution for all h Thus it is impossible to change the entries in b to make Ax b into an inconsistent system No Consider the system as Ax b whereA is a 9X 10 matrix Since the system has a solution for all h in R9 A must have a pivot in each row and so rankA 9 By the Rank Theorem dimNulA 10 9 1 Thus it is impossible to nd two linearly independent vectors in NulA No Consider the system as Ax 0 whereA is a lOgtltl2 matrix SinceA has at most 10 pivot positions rankA S 10 By the Rank Theorem dimNulA 12 rankA 2 2 Thus it is impossible to find a single vector in NulA which spans NulA Yes six equations are sufficient Consider the system as Ax 0 whereA is a l2gtlt8 matrix The problem states that dimNul A 2 By the Rank Theorem rank A 8 dimNul A 6 Thus dimCol A rank A 6 So the system Ax 0 is equivalent to the system Bx 0 where B is an echelon form of A with 6 nonzero rows So the six equations in this system are sufficient to describe the solution set of Ax 0 Yes No Consider the system as Ax b whereA is a 7x6 matrix SinceA has at most 6 pivot positions rank A S 6 By the Rank Theorem dim Nul A 6 rank A 2 0 If dimNul A 0 then the system Ax b will have no free variables The solution to Ax b if it exists would thus have to be unique Since rankA S 6 ColA will be a proper subspace of R7 Thus there exists a bin R7 for which the system AX b is inconsistent and the system Ax b cannot have a unique solution for all h No Consider the system as Ax b whereA is a lOgtltl2 matrix The problem states that dim NulA 3 By the Rank Theorem dimCol A rank A 12 dimNul A 9 Thus ColA will be a proper subspace of R10 Thus there exists a b in R10 for which the system Ax b is inconsistent and the system Ax b cannot have a solution for all h Consider the system Ax 0 whereA is a mxn matrix with m gt n Since the rank ofA is the number of pivot positions thatA has andA is assumed to have full rank rankA n By the Rank Theorem dimNulA n rankA 0 So NulA 0 and the system Ax 0 has only the trivial solution This happens if and only if the columns of A are linearly independent SinceA is an m X n matrix RowA is a subspace ofRquot ColA is a subspace of R and NulA is a subspace of Rquot Likewise since AT is an n X m matrix RowAT is a subspace ome ColAT is a DJ p A Let A 111 46 Solutions 225 subspace of Rquot and NulAT is a subspace ofR39quot Since RowA ColAT and ColA Row AT there are four dinstict subspaces in the list Row A ColA NulA and NulAT a SinceA is an m X n matrix and dimRowA rankA dimRowA dimNulA rankA dimNulA n b Since AT is an n X m matrix and dimColA dimRow A dimColAT rank AT dimCol A dimNul AT rank AT dimNul AT m LetA be an m X n matrix The systemAx b will have a solution for all h in R ifand only ifA has a pivot position in each row which happens if and only if dimColA m By Exercise 28 b dimColA m if and only if dimNulAT m m 0 or NulAT 0 Finally NulAT 0 if and only if the equation ATX 0 has only the trivial solution The equationAx b is consistent if and only if rank A b rank A because the two ranks will be equal if and only if b is not a pivot column of A b The result then follows from Theorem 2 in Section 12 2 2a 2b 2c Compute that uvT 3 a b c 3a 3b 3c Each column of uvT is a multiple of u so 5 5a 5b Sc dimCol uvT 1 unless a b c 0 in which case uvT is the 3 X 3 zero matrix and dimCol uvT 0 In any case rank uvT dimCol uvT S l Note that the second row of the matrix is twice the first row Thus if v l 73 4 which is the first row of the matrix uvT Hg 3 4 21 3 4 6 839 U2 U3 and assume that rankA 1 Suppose that 111 0 Then 111 is basis for ColA since ColA is assumed to be onedimensional Thus there are scalars x and y with 112 xu1 and l 113 yu1 and Au1vT where v x 3 If u1 0 but u2 0 then similarly uz is basis for ColA since ColA is assumed to be one 0 dimensional Thus there is a scalarxwith u3 xuz and A usz where v l x 0 If u1 u2 0 but 113 0 then Au3vT where v 0 l LetA be an m X n matrix with ofrank r gt 0 and let Ube an echelon form ofA SinceA can be reduced to U by row operations there exist invertible elementary matrices E M E P with E P mElA U Thus 226 CHAPTER4 Vector Spaces A E P E1 1U since the product of invertible matrices is invertible Let E E P u E1 1 then A EU Let the columns ofE be denoted by cl cm Since the rank ofA is r Uhas r nonzero rows which can be denoted le wad By the columnrow expansion of A Theorem 10 in Section 24 df T AEUc1 cmd cld1Tmcrdf 0 0 which is the sum of r rank 1 matrices 35 M a Begin by reducingA to reduced echelon form 7 0 13 2 0 5 0 3 0 1 11 2 0 1 2 0 2 A 0 0 0 1 11 2 0 7 0 0 0 0 0 1 1 70 0 0 0 0 0 0 A basis for ColA is the pivot columns of A so matrix C contains these columns 7 7 9 5 3 4 6 2 5 C 5 7 5 2 3 5 l 4 7 6 8 4 9 A basis for Row A is the pivot rows of the reduced echelon form ofA so matrix R contains these rows 71 0 13 2 0 5 0 3 0 1 11 2 0 l 2 0 2 R 0 0 0 1 11 2 0 7 39 70 0 0 0 0 1 1 To nd a basis for NulA row reduce to reduced echelon form note that the solution to Ax 0 in terms of free variables is x1 l32x3 5x5 3x7 x2 11 2x3 l 2x5 2x7 x4 ll2x5 7x7 x6 x7 with x3 x5 and x7 free Thus matrixNis 132 5 3 ll2 l2 2 l 0 0 N 0 112 7 0 1 0 0 0 1 0 0 1 36 M HA is nonzero thenA CR Note that CR Cr1 46 Solutions 227 b The reduced echelon form of AT is 2 11 41 ll 0 2811 0 0 0 I K 2 O O O O O O p O O O O O t O O O O p O O O O p O 0 so the solution to ATX 0 in terms offree variables is x1 211x5 x2 4111x5 x3 0 x4 2811x5 with x5 free Thus matrixM is 2 1 1 41 11 M 0 28 1 1 1 The matrix S RT N J is 7 X 7 because the columns of RT andN are in R7 and dimRowA dimNulA 7 The matrix T C M is 5 X 5 because the columns ofC andMare in R5 and dimColA dimNul AT 5 Both S and T are invertible because their columns are linearly independent This fact will be proven in general in Theorem 3 of Section 61 M Answers will vary but in most cases C will be 6 X 4 and will be constructed from the rst 4 columns ofA In most cases R will be 4 X 7 Nwill be 7 X 3 andeill be 6 X 2 M The C and R from Exercise 35 work here andA CR M r are the Cr2 columns of R The columns of R are either pivot columns of R or are not pivot columns of R Cr where r1 Consider rst the pivot columns of R The in pivot column of R is e1 the 1quot11 column in the identity matrix so Cel is the in pivot column of A SinceA and R have pivot columns in the same locations when C multiplies a pivot column of R the result is the corresponding pivot column ofA in its proper location Suppose r J is a nonpivot column of R Then r J contains the weights needed to construct the 1quotquot column of A from the pivot columns ofA as is discussed in Example 9 of Section 43 and in the paragraph preceding that example Thus r J contains the weights needed to construct the 1quot column of A from the columns of C and Cr a 228 CHAPTER4 Vector Spaces 47 SOLUTIONS Notes This section depends heavily on the coordinate systems introduced in Section 44 The row reduction algorithm that produces 18 can also be deduced from Exercise 12 in Section 22 by row reducing PC IPB to I IPCTIPB The changeof coordinates matrix here is interpreted in Section 54 as the matrix of the identity transformation relative to two bases 6 9 6 9 1 21 Since b1 601 202 and b2 901 402 b1C 2 b2C 4 and C58 2 4 3 b Since x 3bl2b2 XB 2 and 6 9 3 0 XCC1Bxlg 2 4 2 2 l 5 1 5 2 21 Since b1 cl4c2 and b2 501 3cz b1C 4 b2C 3 and C58 4 g 5 b Since x5bl3b2 XB 3 and 1 5 5 10 Allin 3 Equation ii is satis ed by P for all x in V 4 Equation i is satis ed by P for all x in V 4 l 539 339 Since alz4blb2 a2bib2b3a and a3122b3 brig 1 aZB 1 0 l 0 4 l 0 21318 1andBIA 1 1 2 0 l 2 3 b Since x3a14a2a3 xA 4 and l 4 l 0 3 8 XBBIA l l l 4 2 l 2 l 2 47 Solutions 229 2 0 3 6 21 Since f12d1 d2d3 f23d2d3and f3 3d12d3 f1D 1f2D 3 f3D 0 1 1 2 2 0 3 and P 1 3 0 DFF 1 1 2 1 b Since xf1 2f22f3 xF 2 and 2 2 0 3 1 4 xDD1FxF 1 3 0 2 7 1 1 2 2 3 7 To nd CPR row reduce the matrix c1 c2 b1 b2 c1 c2 b1 b2 10 31 01 5239 3 1 1 2 1 Thus P and P P C93 5 2 39C C93 5 3 8 To nd P row reduce the matrix c1 c2 b1 b2 C98 1 0 3 2 cl c2 b1 b2 0 1 4 3 3 2 3 2 Thus P and P P 71 C93 4 3 39C C93 4 3 9 To nd P row reduce the matrix c1 c2 b1 b2 C98 1 0 9 2 cl c2 b1 b2 0 1 4 1 9 2 1 1 2 Thus P and P P C93 4 1 39C C93 4 9 10 To nd P row reduce the matrix c1 c2 b1 b2 C98 10 8 3 cl c2 b1 b2 0 1 5 2 8 3 2 3 Thus P 2 and P P71 C93 5 39C C93 5 8 11 21 False See Theorem 15 b True See the rst paragraph in the subsection Change of Basis in Rquot 230 CHAPTER4 Vector Spaces 12 a True The columns of I are coordinate vectors of the linearly independent set B See the second C B paragraph after Theorem 15 b False The row reduction is discussed after Example 2 The matrix P obtained there satis es ch Plxlg 13 Let Bb1b2b31 2tt23 5t4t2 2t3t2 and let Cclczc3l M The C coordinate vectors of b1 b2 and b3 are l 3 0 b1C 2 b2C 5 b3lcz 2 l 4 3 So 1 3 0 2 5 2 CltB l 4 3 Let x 71 21 Then the coordinate vector XB satis es 1 CEBMB x1c 2 0 This system may be solved by row reducing its augmented matrix 1 3 0 l l 0 0 5 5 2 5 2 20 1 0 2soxB 2 1 4 3 0 0 0 0 1 1 14 Let B b1 b2 b3 1 31 2 2 t 51 2 1 21 and let C 01 cz c3 1 1 t2 The Ccoordinate vectors of b1 b2 and b3 are l 2 1 b1lc 0 bzlc 1 b3lc 2 3 5 0 So 1 2 l P 0 1 2 C B 3 5 0 Let x t2 Then the coordinate vector XB satisfies 0 CElelg ch 0 l 47 Solutions 231 This system may be solved by row reducing its augmented matrix 1210 100 3 3 0 12 0010 2s0xB 2 3 501 0001 1 and 2 31 3t2 22t 5t212t Equot a B is a basis for V b the coordinate mapping is a linear transformation c of the product of a matrix and a vector d the coordinate vector of v relative to B 1 16 a b1C Qb1B Q 2 Qe1 3 b bkC C bkC QbkB Q91 17 M 21 Since we found P in Exercise 34 of Section 45 we can calculate that 32 0 16 0 12 0 10 0 32 0 24 0 20 0 1 0 0 16 0 16 0 15 P 1 0 0 0 8 0 10 0 32 0 0 0 0 4 0 6 0 0 0 0 0 2 0 0 0 0 0 0 0 1 b Since P is the changeofcoordinates matrix from C to B P71 will be the changeof coordinates matrix fromB to C By Theorem 15 the columns of P 1 will be the Ccoordinate vectors of the basis vectors in B Thus coszt 1 cos 2t 3 1 cos I 23c0s t cos 3t cos4t 3 4cos 21 cos 4t cosst Ocos t 5 cos 31 cos 5t cos6t3 121015cos 2t6cos 4tcos 6t

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